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Existence of Solution, System of Equations
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Suppose $P(lambda, i)$ is the probability that a Poisson random variable with average $lambda$ is equal to $i$, i.e. $fraclambda^ie^lambdai!$
I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $alpha>0$ and $kin mathbbN_+$
begincases
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i) \
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i+1)
endcases
where the necessary condition $alphaleq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $alpha=P(lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $lambda$, the largest solution to the equation $alpha=P(lambda,k+1)$. Experiments show that for each fixed $alpha$ and $k$, there is a solution, but I did not manage to prove it analytically.
Is there any analogue for mean value theorem for multidimensional functions?
Any suggestion for the proof directions will be appreciated.
fa.functional-analysis pr.probability real-analysis fixed-point-theorems functional-equations
asked 12 hours ago
Tina
283
add a comment |Â
up vote
5
down vote
favorite
Suppose $P(lambda, i)$ is the probability that a Poisson random variable with average $lambda$ is equal to $i$, i.e. $fraclambda^ie^lambdai!$
I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $alpha>0$ and $kin mathbbN_+$
begincases
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i) \
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i+1)
endcases
where the necessary condition $alphaleq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $alpha=P(lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $lambda$, the largest solution to the equation $alpha=P(lambda,k+1)$. Experiments show that for each fixed $alpha$ and $k$, there is a solution, but I did not manage to prove it analytically.
Is there any analogue for mean value theorem for multidimensional functions?
Any suggestion for the proof directions will be appreciated.
fa.functional-analysis pr.probability real-analysis fixed-point-theorems functional-equations
asked 12 hours ago
Tina
283
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose $P(lambda, i)$ is the probability that a Poisson random variable with average $lambda$ is equal to $i$, i.e. $fraclambda^ie^lambdai!$
I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $alpha>0$ and $kin mathbbN_+$
begincases
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i) \
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i+1)
endcases
where the necessary condition $alphaleq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $alpha=P(lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $lambda$, the largest solution to the equation $alpha=P(lambda,k+1)$. Experiments show that for each fixed $alpha$ and $k$, there is a solution, but I did not manage to prove it analytically.
Is there any analogue for mean value theorem for multidimensional functions?
Any suggestion for the proof directions will be appreciated.
fa.functional-analysis pr.probability real-analysis fixed-point-theorems functional-equations
asked 12 hours ago
Tina
283
Suppose $P(lambda, i)$ is the probability that a Poisson random variable with average $lambda$ is equal to $i$, i.e. $fraclambda^ie^lambdai!$
I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $alpha>0$ and $kin mathbbN_+$
begincases
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i) \
alpha=sum_i=0^inftyP(x, i)cdot P(y, k+i+1)
endcases
where the necessary condition $alphaleq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $alpha=P(lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $lambda$, the largest solution to the equation $alpha=P(lambda,k+1)$. Experiments show that for each fixed $alpha$ and $k$, there is a solution, but I did not manage to prove it analytically.
Is there any analogue for mean value theorem for multidimensional functions?
Any suggestion for the proof directions will be appreciated.
fa.functional-analysis pr.probability real-analysis fixed-point-theorems functional-equations
asked 12 hours ago
Tina
283
asked 12 hours ago
Tina
283
asked 12 hours ago
Tina
283
asked 12 hours ago
Tina
283
283
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Let $a:=alpha$ and
beginequation*
F_k(x,y):=sum_j=0^infty fracx^jj!fracy^k+j(k+j)!,e^-x-y,
endequation*
assuming the standard convention $0^0:=1$.
We have to consider the existence of a solution in $x$ and $y$ of the system
beginequation*
a=F_k(x,y)=F_k+1(x,y). tag1
endequation*
We shall prove the following.
Theorem 1. Take any natural $k$ and any
beginequation*
ain(0,a_k],quadtextwherequad a_k:=sup_x,yge0F_k+1(x,y). tag1.5
endequation*
Then the system (1) has a solution $x,yge0$.
Remark 1. Since $F_k>0$, the condition $ain(0,a_k]$ is obviously necessary in Theorem 1.
Proof of Theorem 1. Note that $F_k(x,y)ge0$ for any real $x,yge0$ and $F_k(x,y)$ is continuous in real $x,yge0$. The crucial observation is the identity
beginequation*
partial_y F_k+1(x,y)=F_k(x,y)-F_k+1(x,y) tag2
endequation*
for real $x,y$.
Next, fix for a moment any real $xge0$. Then $F_k+1(x,0)=0$ and, by dominated convergence, $F_k+1(x,infty-)=0$. So, $F_k+1(x,y)$ attains its maximum in $y$ at some real point $y=y_xge0$. At this point, we have $partial_y F_k+1(x,y)=0$. So, by (2),
beginequation*
F_k(x,y_x)=F_k+1(x,y_x)=max_yge0F_k+1(x,y)=:M_k+1(x), tag3
endequation*
for all real $xge0$.
Next,
beginalign*
M_k(x)&lesum_j=0^infty fracx^jj!max_yge0fracy^k+j(k+j)!,e^-x-y \
&=sum_j=0^infty fracx^jj!e^-xfrac(k+j)^k+j(k+j)!,e^-k-j tag4 \
&llsum_j=0^infty fracx^jj!e^-xfrac1sqrtk+j=Efrac1sqrtk+Pi_x
undersetxtoinftylongrightarrow0
endalign*
by dominated convergence and because $Pi_xundersetxtoinftylongrightarrowinfty$ in probability, where $Pi_x$ is a Poisson random variable with parameter $x$. So,
$M_k(infty-)=0$. It is also not hard to see that
$F_k(x,y)$ is continuous in real $xge0$ uniformly in real $yge0$ (see the Appendix), so that $M_k(x)$ is continuous in $xge0$. So, $M_k+1(x)$ attains its maximum in $xge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $qquadBox$
Appendix. Similarly to (2),
beginequation*
partial_x F_k(x,y)=F_k+1(x,y)-F_k(x,y).
endequation*
for real $x,y$. Therefore and because $0le F_kle1$, we have $|partial_x F_k(x,y)|le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $xge0$ uniformly in real $yge0$.
Added: Let us now show that
beginequation*
a_k=c_k+1,quadtextwherequad c_k:=frack^kk!,e^-k
simfrac1sqrt2pi k
endequation*
as $ktoinfty$. To this end, note first that
beginequation*
c_k+1/c_k=(1+1/k)^k/e<1,
endequation*
and so, $c_k$ is decreasing in $k$.
So, recalling (4), we have
beginequation*
M_k(x)lesum_j=0^infty fracx^jj!e^-x,c_k+j
lesum_j=0^infty fracx^jj!e^-x,c_k=c_k=M_k(0).
endequation*
Thus, in view of (1.5) and (2),
beginequation*
a_k=max_xge0M_k+1(x)=M_k+1(0)=c_k+1,
endequation*
as desired.
In particular, for $k=0,1,2,3$ the values of $a_k$ are $approx0.367879, 0.270671, 0.224042$.
answered 8 hours ago
Iosif Pinelis
14.1k12154
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
 |Â
show 1 more comment
up vote
2
down vote
The sums over $P(lambda,i)=fraclambda^ie^lambdai!$ are evaluated in terms of a Bessel function as
$$sum_i=0^inftyP(x, i)cdot P(y, k+i)=y^k e^-x-y left(xyright)^-k/2 I_kleft(2 sqrtxyright)$$
$$sum_i=0^inftyP(x, i)cdot P(y, k+i+1)=(y/x)^1/2,y^k e^-x-y left(xyright)^-k/2 I_k+1left(2 sqrtxyright)$$
for any positive integer $k$ these two expressions should be equal for some $x,y>0$. (For $x=y=0$ both expressions are identically zero.) So the function
$$F_k(x,y)=sqrtx, I_kleft(2 sqrtxyright)-sqrty ,I_k+1left(2 sqrtxyright)$$
should pass through zero in the quadrant $x,y>0$ for any positive integer $k$.
For large $z=2sqrtxy$ both Bessel functions $I_k(z)$ and $I_k+1(z)$ grow as $(2pi z)^-1/2e^z$, so by making $x$ much larger than $y$ the function $F_k(x,y)$ is positive and by making $y$ much larger than $x$ it is negative, hence it must go through zero when $xapprox y$.
I had not appreciated that $alpha$ is fixed from the beginning like $k$, not a variable like $x$ and $y$. So we also need to show that $xapprox ygg 1$ allows the sum to reach any $alpha>0$, so
$$alpha=e^-2x I_kleft(2xright)approx (4pi x)^-1/2,;;xgg 1.$$
This is possible only for $alphall 1$. The OP lists as necessary condition $alphaleq P(k+1,k+1)$, it is not clear to me this is sufficient.
answered 11 hours ago
Carlo Beenakker
67.3k6153252
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
2
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $a:=alpha$ and
beginequation*
F_k(x,y):=sum_j=0^infty fracx^jj!fracy^k+j(k+j)!,e^-x-y,
endequation*
assuming the standard convention $0^0:=1$.
We have to consider the existence of a solution in $x$ and $y$ of the system
beginequation*
a=F_k(x,y)=F_k+1(x,y). tag1
endequation*
We shall prove the following.
Theorem 1. Take any natural $k$ and any
beginequation*
ain(0,a_k],quadtextwherequad a_k:=sup_x,yge0F_k+1(x,y). tag1.5
endequation*
Then the system (1) has a solution $x,yge0$.
Remark 1. Since $F_k>0$, the condition $ain(0,a_k]$ is obviously necessary in Theorem 1.
Proof of Theorem 1. Note that $F_k(x,y)ge0$ for any real $x,yge0$ and $F_k(x,y)$ is continuous in real $x,yge0$. The crucial observation is the identity
beginequation*
partial_y F_k+1(x,y)=F_k(x,y)-F_k+1(x,y) tag2
endequation*
for real $x,y$.
Next, fix for a moment any real $xge0$. Then $F_k+1(x,0)=0$ and, by dominated convergence, $F_k+1(x,infty-)=0$. So, $F_k+1(x,y)$ attains its maximum in $y$ at some real point $y=y_xge0$. At this point, we have $partial_y F_k+1(x,y)=0$. So, by (2),
beginequation*
F_k(x,y_x)=F_k+1(x,y_x)=max_yge0F_k+1(x,y)=:M_k+1(x), tag3
endequation*
for all real $xge0$.
Next,
beginalign*
M_k(x)&lesum_j=0^infty fracx^jj!max_yge0fracy^k+j(k+j)!,e^-x-y \
&=sum_j=0^infty fracx^jj!e^-xfrac(k+j)^k+j(k+j)!,e^-k-j tag4 \
&llsum_j=0^infty fracx^jj!e^-xfrac1sqrtk+j=Efrac1sqrtk+Pi_x
undersetxtoinftylongrightarrow0
endalign*
by dominated convergence and because $Pi_xundersetxtoinftylongrightarrowinfty$ in probability, where $Pi_x$ is a Poisson random variable with parameter $x$. So,
$M_k(infty-)=0$. It is also not hard to see that
$F_k(x,y)$ is continuous in real $xge0$ uniformly in real $yge0$ (see the Appendix), so that $M_k(x)$ is continuous in $xge0$. So, $M_k+1(x)$ attains its maximum in $xge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $qquadBox$
Appendix. Similarly to (2),
beginequation*
partial_x F_k(x,y)=F_k+1(x,y)-F_k(x,y).
endequation*
for real $x,y$. Therefore and because $0le F_kle1$, we have $|partial_x F_k(x,y)|le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $xge0$ uniformly in real $yge0$.
Added: Let us now show that
beginequation*
a_k=c_k+1,quadtextwherequad c_k:=frack^kk!,e^-k
simfrac1sqrt2pi k
endequation*
as $ktoinfty$. To this end, note first that
beginequation*
c_k+1/c_k=(1+1/k)^k/e<1,
endequation*
and so, $c_k$ is decreasing in $k$.
So, recalling (4), we have
beginequation*
M_k(x)lesum_j=0^infty fracx^jj!e^-x,c_k+j
lesum_j=0^infty fracx^jj!e^-x,c_k=c_k=M_k(0).
endequation*
Thus, in view of (1.5) and (2),
beginequation*
a_k=max_xge0M_k+1(x)=M_k+1(0)=c_k+1,
endequation*
as desired.
In particular, for $k=0,1,2,3$ the values of $a_k$ are $approx0.367879, 0.270671, 0.224042$.
answered 8 hours ago
Iosif Pinelis
14.1k12154
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
 |Â
show 1 more comment
up vote
4
down vote
accepted
Let $a:=alpha$ and
beginequation*
F_k(x,y):=sum_j=0^infty fracx^jj!fracy^k+j(k+j)!,e^-x-y,
endequation*
assuming the standard convention $0^0:=1$.
We have to consider the existence of a solution in $x$ and $y$ of the system
beginequation*
a=F_k(x,y)=F_k+1(x,y). tag1
endequation*
We shall prove the following.
Theorem 1. Take any natural $k$ and any
beginequation*
ain(0,a_k],quadtextwherequad a_k:=sup_x,yge0F_k+1(x,y). tag1.5
endequation*
Then the system (1) has a solution $x,yge0$.
Remark 1. Since $F_k>0$, the condition $ain(0,a_k]$ is obviously necessary in Theorem 1.
Proof of Theorem 1. Note that $F_k(x,y)ge0$ for any real $x,yge0$ and $F_k(x,y)$ is continuous in real $x,yge0$. The crucial observation is the identity
beginequation*
partial_y F_k+1(x,y)=F_k(x,y)-F_k+1(x,y) tag2
endequation*
for real $x,y$.
Next, fix for a moment any real $xge0$. Then $F_k+1(x,0)=0$ and, by dominated convergence, $F_k+1(x,infty-)=0$. So, $F_k+1(x,y)$ attains its maximum in $y$ at some real point $y=y_xge0$. At this point, we have $partial_y F_k+1(x,y)=0$. So, by (2),
beginequation*
F_k(x,y_x)=F_k+1(x,y_x)=max_yge0F_k+1(x,y)=:M_k+1(x), tag3
endequation*
for all real $xge0$.
Next,
beginalign*
M_k(x)&lesum_j=0^infty fracx^jj!max_yge0fracy^k+j(k+j)!,e^-x-y \
&=sum_j=0^infty fracx^jj!e^-xfrac(k+j)^k+j(k+j)!,e^-k-j tag4 \
&llsum_j=0^infty fracx^jj!e^-xfrac1sqrtk+j=Efrac1sqrtk+Pi_x
undersetxtoinftylongrightarrow0
endalign*
by dominated convergence and because $Pi_xundersetxtoinftylongrightarrowinfty$ in probability, where $Pi_x$ is a Poisson random variable with parameter $x$. So,
$M_k(infty-)=0$. It is also not hard to see that
$F_k(x,y)$ is continuous in real $xge0$ uniformly in real $yge0$ (see the Appendix), so that $M_k(x)$ is continuous in $xge0$. So, $M_k+1(x)$ attains its maximum in $xge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $qquadBox$
Appendix. Similarly to (2),
beginequation*
partial_x F_k(x,y)=F_k+1(x,y)-F_k(x,y).
endequation*
for real $x,y$. Therefore and because $0le F_kle1$, we have $|partial_x F_k(x,y)|le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $xge0$ uniformly in real $yge0$.
Added: Let us now show that
beginequation*
a_k=c_k+1,quadtextwherequad c_k:=frack^kk!,e^-k
simfrac1sqrt2pi k
endequation*
as $ktoinfty$. To this end, note first that
beginequation*
c_k+1/c_k=(1+1/k)^k/e<1,
endequation*
and so, $c_k$ is decreasing in $k$.
So, recalling (4), we have
beginequation*
M_k(x)lesum_j=0^infty fracx^jj!e^-x,c_k+j
lesum_j=0^infty fracx^jj!e^-x,c_k=c_k=M_k(0).
endequation*
Thus, in view of (1.5) and (2),
beginequation*
a_k=max_xge0M_k+1(x)=M_k+1(0)=c_k+1,
endequation*
as desired.
In particular, for $k=0,1,2,3$ the values of $a_k$ are $approx0.367879, 0.270671, 0.224042$.
answered 8 hours ago
Iosif Pinelis
14.1k12154
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $a:=alpha$ and
beginequation*
F_k(x,y):=sum_j=0^infty fracx^jj!fracy^k+j(k+j)!,e^-x-y,
endequation*
assuming the standard convention $0^0:=1$.
We have to consider the existence of a solution in $x$ and $y$ of the system
beginequation*
a=F_k(x,y)=F_k+1(x,y). tag1
endequation*
We shall prove the following.
Theorem 1. Take any natural $k$ and any
beginequation*
ain(0,a_k],quadtextwherequad a_k:=sup_x,yge0F_k+1(x,y). tag1.5
endequation*
Then the system (1) has a solution $x,yge0$.
Remark 1. Since $F_k>0$, the condition $ain(0,a_k]$ is obviously necessary in Theorem 1.
Proof of Theorem 1. Note that $F_k(x,y)ge0$ for any real $x,yge0$ and $F_k(x,y)$ is continuous in real $x,yge0$. The crucial observation is the identity
beginequation*
partial_y F_k+1(x,y)=F_k(x,y)-F_k+1(x,y) tag2
endequation*
for real $x,y$.
Next, fix for a moment any real $xge0$. Then $F_k+1(x,0)=0$ and, by dominated convergence, $F_k+1(x,infty-)=0$. So, $F_k+1(x,y)$ attains its maximum in $y$ at some real point $y=y_xge0$. At this point, we have $partial_y F_k+1(x,y)=0$. So, by (2),
beginequation*
F_k(x,y_x)=F_k+1(x,y_x)=max_yge0F_k+1(x,y)=:M_k+1(x), tag3
endequation*
for all real $xge0$.
Next,
beginalign*
M_k(x)&lesum_j=0^infty fracx^jj!max_yge0fracy^k+j(k+j)!,e^-x-y \
&=sum_j=0^infty fracx^jj!e^-xfrac(k+j)^k+j(k+j)!,e^-k-j tag4 \
&llsum_j=0^infty fracx^jj!e^-xfrac1sqrtk+j=Efrac1sqrtk+Pi_x
undersetxtoinftylongrightarrow0
endalign*
by dominated convergence and because $Pi_xundersetxtoinftylongrightarrowinfty$ in probability, where $Pi_x$ is a Poisson random variable with parameter $x$. So,
$M_k(infty-)=0$. It is also not hard to see that
$F_k(x,y)$ is continuous in real $xge0$ uniformly in real $yge0$ (see the Appendix), so that $M_k(x)$ is continuous in $xge0$. So, $M_k+1(x)$ attains its maximum in $xge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $qquadBox$
Appendix. Similarly to (2),
beginequation*
partial_x F_k(x,y)=F_k+1(x,y)-F_k(x,y).
endequation*
for real $x,y$. Therefore and because $0le F_kle1$, we have $|partial_x F_k(x,y)|le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $xge0$ uniformly in real $yge0$.
Added: Let us now show that
beginequation*
a_k=c_k+1,quadtextwherequad c_k:=frack^kk!,e^-k
simfrac1sqrt2pi k
endequation*
as $ktoinfty$. To this end, note first that
beginequation*
c_k+1/c_k=(1+1/k)^k/e<1,
endequation*
and so, $c_k$ is decreasing in $k$.
So, recalling (4), we have
beginequation*
M_k(x)lesum_j=0^infty fracx^jj!e^-x,c_k+j
lesum_j=0^infty fracx^jj!e^-x,c_k=c_k=M_k(0).
endequation*
Thus, in view of (1.5) and (2),
beginequation*
a_k=max_xge0M_k+1(x)=M_k+1(0)=c_k+1,
endequation*
as desired.
In particular, for $k=0,1,2,3$ the values of $a_k$ are $approx0.367879, 0.270671, 0.224042$.
answered 8 hours ago
Iosif Pinelis
14.1k12154
Let $a:=alpha$ and
beginequation*
F_k(x,y):=sum_j=0^infty fracx^jj!fracy^k+j(k+j)!,e^-x-y,
endequation*
assuming the standard convention $0^0:=1$.
We have to consider the existence of a solution in $x$ and $y$ of the system
beginequation*
a=F_k(x,y)=F_k+1(x,y). tag1
endequation*
We shall prove the following.
Theorem 1. Take any natural $k$ and any
beginequation*
ain(0,a_k],quadtextwherequad a_k:=sup_x,yge0F_k+1(x,y). tag1.5
endequation*
Then the system (1) has a solution $x,yge0$.
Remark 1. Since $F_k>0$, the condition $ain(0,a_k]$ is obviously necessary in Theorem 1.
Proof of Theorem 1. Note that $F_k(x,y)ge0$ for any real $x,yge0$ and $F_k(x,y)$ is continuous in real $x,yge0$. The crucial observation is the identity
beginequation*
partial_y F_k+1(x,y)=F_k(x,y)-F_k+1(x,y) tag2
endequation*
for real $x,y$.
Next, fix for a moment any real $xge0$. Then $F_k+1(x,0)=0$ and, by dominated convergence, $F_k+1(x,infty-)=0$. So, $F_k+1(x,y)$ attains its maximum in $y$ at some real point $y=y_xge0$. At this point, we have $partial_y F_k+1(x,y)=0$. So, by (2),
beginequation*
F_k(x,y_x)=F_k+1(x,y_x)=max_yge0F_k+1(x,y)=:M_k+1(x), tag3
endequation*
for all real $xge0$.
Next,
beginalign*
M_k(x)&lesum_j=0^infty fracx^jj!max_yge0fracy^k+j(k+j)!,e^-x-y \
&=sum_j=0^infty fracx^jj!e^-xfrac(k+j)^k+j(k+j)!,e^-k-j tag4 \
&llsum_j=0^infty fracx^jj!e^-xfrac1sqrtk+j=Efrac1sqrtk+Pi_x
undersetxtoinftylongrightarrow0
endalign*
by dominated convergence and because $Pi_xundersetxtoinftylongrightarrowinfty$ in probability, where $Pi_x$ is a Poisson random variable with parameter $x$. So,
$M_k(infty-)=0$. It is also not hard to see that
$F_k(x,y)$ is continuous in real $xge0$ uniformly in real $yge0$ (see the Appendix), so that $M_k(x)$ is continuous in $xge0$. So, $M_k+1(x)$ attains its maximum in $xge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $qquadBox$
Appendix. Similarly to (2),
beginequation*
partial_x F_k(x,y)=F_k+1(x,y)-F_k(x,y).
endequation*
for real $x,y$. Therefore and because $0le F_kle1$, we have $|partial_x F_k(x,y)|le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $xge0$ uniformly in real $yge0$.
Added: Let us now show that
beginequation*
a_k=c_k+1,quadtextwherequad c_k:=frack^kk!,e^-k
simfrac1sqrt2pi k
endequation*
as $ktoinfty$. To this end, note first that
beginequation*
c_k+1/c_k=(1+1/k)^k/e<1,
endequation*
and so, $c_k$ is decreasing in $k$.
So, recalling (4), we have
beginequation*
M_k(x)lesum_j=0^infty fracx^jj!e^-x,c_k+j
lesum_j=0^infty fracx^jj!e^-x,c_k=c_k=M_k(0).
endequation*
Thus, in view of (1.5) and (2),
beginequation*
a_k=max_xge0M_k+1(x)=M_k+1(0)=c_k+1,
endequation*
as desired.
In particular, for $k=0,1,2,3$ the values of $a_k$ are $approx0.367879, 0.270671, 0.224042$.
answered 8 hours ago
Iosif Pinelis
14.1k12154
edited 5 hours ago
answered 8 hours ago
Iosif Pinelis
14.1k12154
answered 8 hours ago
Iosif Pinelis
14.1k12154
answered 8 hours ago
Iosif Pinelis
14.1k12154
14.1k12154
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
 |Â
show 1 more comment
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
I have added Remark 2 concerning the values of $a_k$.
– Iosif Pinelis
7 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
The proof of $a_k=P(k+1,k+1)$ is very easy, if $alpha>P(k+1,k+1)$ then it implies $alpha>P(lambda, k+1)$ for each $lambda$, since $P(lambda,k+1)$ is maximized for $lambda=k+1$. On the other hand, by induction we can prove that $alpha>P(lambda,k+1+i)$ for any $i>0$ and $lambda$, therefore the second equation does not have a solution!
– Tina
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
@Tina : I am afraid I don't understand your comment: (i) So, what if $alpha>P(lambda,k+1)$? What does this imply? (ii) What is your $alpha$ in the inequality $alpha>P(lambda,k+1+i)$? (iii) How do you prove this inequality? (iv) What does this latter inequality imply?
– Iosif Pinelis
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
(i) if $alpha>P(lambda, k+1)$ for any $lambda$ this implies that $alpha>P(lambda,k+2)$ or for any $i>0$, $alpha>P(lambda,k+1+i)$. (ii) $alpha$ is the parameter from the system of equations. (iii) proof is by induction, if $alpha>P(lambda,k+1)$, this means $alphacdot e^lambda - fraclambda^k+1(k+1)!>0$, then $alphacdot e^lambda - fraclambda^k+2(k+2)!>0$, since the latter holds for $lambda=0$ and its derivative is the former, always positive. (iv) this implies that the second inequality of the system is not solvable, since $sum_i=0^inftyP(x,i)=1$.
– Tina
6 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
@Tina : I am sorry, I still don't understand your comment.
– Iosif Pinelis
5 hours ago
 |Â
show 1 more comment
up vote
2
down vote
The sums over $P(lambda,i)=fraclambda^ie^lambdai!$ are evaluated in terms of a Bessel function as
$$sum_i=0^inftyP(x, i)cdot P(y, k+i)=y^k e^-x-y left(xyright)^-k/2 I_kleft(2 sqrtxyright)$$
$$sum_i=0^inftyP(x, i)cdot P(y, k+i+1)=(y/x)^1/2,y^k e^-x-y left(xyright)^-k/2 I_k+1left(2 sqrtxyright)$$
for any positive integer $k$ these two expressions should be equal for some $x,y>0$. (For $x=y=0$ both expressions are identically zero.) So the function
$$F_k(x,y)=sqrtx, I_kleft(2 sqrtxyright)-sqrty ,I_k+1left(2 sqrtxyright)$$
should pass through zero in the quadrant $x,y>0$ for any positive integer $k$.
For large $z=2sqrtxy$ both Bessel functions $I_k(z)$ and $I_k+1(z)$ grow as $(2pi z)^-1/2e^z$, so by making $x$ much larger than $y$ the function $F_k(x,y)$ is positive and by making $y$ much larger than $x$ it is negative, hence it must go through zero when $xapprox y$.
I had not appreciated that $alpha$ is fixed from the beginning like $k$, not a variable like $x$ and $y$. So we also need to show that $xapprox ygg 1$ allows the sum to reach any $alpha>0$, so
$$alpha=e^-2x I_kleft(2xright)approx (4pi x)^-1/2,;;xgg 1.$$
This is possible only for $alphall 1$. The OP lists as necessary condition $alphaleq P(k+1,k+1)$, it is not clear to me this is sufficient.
answered 11 hours ago
Carlo Beenakker
67.3k6153252
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
2
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
add a comment |Â
up vote
2
down vote
The sums over $P(lambda,i)=fraclambda^ie^lambdai!$ are evaluated in terms of a Bessel function as
$$sum_i=0^inftyP(x, i)cdot P(y, k+i)=y^k e^-x-y left(xyright)^-k/2 I_kleft(2 sqrtxyright)$$
$$sum_i=0^inftyP(x, i)cdot P(y, k+i+1)=(y/x)^1/2,y^k e^-x-y left(xyright)^-k/2 I_k+1left(2 sqrtxyright)$$
for any positive integer $k$ these two expressions should be equal for some $x,y>0$. (For $x=y=0$ both expressions are identically zero.) So the function
$$F_k(x,y)=sqrtx, I_kleft(2 sqrtxyright)-sqrty ,I_k+1left(2 sqrtxyright)$$
should pass through zero in the quadrant $x,y>0$ for any positive integer $k$.
For large $z=2sqrtxy$ both Bessel functions $I_k(z)$ and $I_k+1(z)$ grow as $(2pi z)^-1/2e^z$, so by making $x$ much larger than $y$ the function $F_k(x,y)$ is positive and by making $y$ much larger than $x$ it is negative, hence it must go through zero when $xapprox y$.
I had not appreciated that $alpha$ is fixed from the beginning like $k$, not a variable like $x$ and $y$. So we also need to show that $xapprox ygg 1$ allows the sum to reach any $alpha>0$, so
$$alpha=e^-2x I_kleft(2xright)approx (4pi x)^-1/2,;;xgg 1.$$
This is possible only for $alphall 1$. The OP lists as necessary condition $alphaleq P(k+1,k+1)$, it is not clear to me this is sufficient.
answered 11 hours ago
Carlo Beenakker
67.3k6153252
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
2
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The sums over $P(lambda,i)=fraclambda^ie^lambdai!$ are evaluated in terms of a Bessel function as
$$sum_i=0^inftyP(x, i)cdot P(y, k+i)=y^k e^-x-y left(xyright)^-k/2 I_kleft(2 sqrtxyright)$$
$$sum_i=0^inftyP(x, i)cdot P(y, k+i+1)=(y/x)^1/2,y^k e^-x-y left(xyright)^-k/2 I_k+1left(2 sqrtxyright)$$
for any positive integer $k$ these two expressions should be equal for some $x,y>0$. (For $x=y=0$ both expressions are identically zero.) So the function
$$F_k(x,y)=sqrtx, I_kleft(2 sqrtxyright)-sqrty ,I_k+1left(2 sqrtxyright)$$
should pass through zero in the quadrant $x,y>0$ for any positive integer $k$.
For large $z=2sqrtxy$ both Bessel functions $I_k(z)$ and $I_k+1(z)$ grow as $(2pi z)^-1/2e^z$, so by making $x$ much larger than $y$ the function $F_k(x,y)$ is positive and by making $y$ much larger than $x$ it is negative, hence it must go through zero when $xapprox y$.
I had not appreciated that $alpha$ is fixed from the beginning like $k$, not a variable like $x$ and $y$. So we also need to show that $xapprox ygg 1$ allows the sum to reach any $alpha>0$, so
$$alpha=e^-2x I_kleft(2xright)approx (4pi x)^-1/2,;;xgg 1.$$
This is possible only for $alphall 1$. The OP lists as necessary condition $alphaleq P(k+1,k+1)$, it is not clear to me this is sufficient.
answered 11 hours ago
Carlo Beenakker
67.3k6153252
The sums over $P(lambda,i)=fraclambda^ie^lambdai!$ are evaluated in terms of a Bessel function as
$$sum_i=0^inftyP(x, i)cdot P(y, k+i)=y^k e^-x-y left(xyright)^-k/2 I_kleft(2 sqrtxyright)$$
$$sum_i=0^inftyP(x, i)cdot P(y, k+i+1)=(y/x)^1/2,y^k e^-x-y left(xyright)^-k/2 I_k+1left(2 sqrtxyright)$$
for any positive integer $k$ these two expressions should be equal for some $x,y>0$. (For $x=y=0$ both expressions are identically zero.) So the function
$$F_k(x,y)=sqrtx, I_kleft(2 sqrtxyright)-sqrty ,I_k+1left(2 sqrtxyright)$$
should pass through zero in the quadrant $x,y>0$ for any positive integer $k$.
For large $z=2sqrtxy$ both Bessel functions $I_k(z)$ and $I_k+1(z)$ grow as $(2pi z)^-1/2e^z$, so by making $x$ much larger than $y$ the function $F_k(x,y)$ is positive and by making $y$ much larger than $x$ it is negative, hence it must go through zero when $xapprox y$.
I had not appreciated that $alpha$ is fixed from the beginning like $k$, not a variable like $x$ and $y$. So we also need to show that $xapprox ygg 1$ allows the sum to reach any $alpha>0$, so
$$alpha=e^-2x I_kleft(2xright)approx (4pi x)^-1/2,;;xgg 1.$$
This is possible only for $alphall 1$. The OP lists as necessary condition $alphaleq P(k+1,k+1)$, it is not clear to me this is sufficient.
answered 11 hours ago
Carlo Beenakker
67.3k6153252
edited 9 hours ago
answered 11 hours ago
Carlo Beenakker
67.3k6153252
answered 11 hours ago
Carlo Beenakker
67.3k6153252
answered 11 hours ago
Carlo Beenakker
67.3k6153252
67.3k6153252
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
2
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
add a comment |Â
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
2
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
Thank you! But I do not understand why both terms are equal to $alpha$?
– Tina
11 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
ah wait, $alpha$ is fixed from the beginning like $k$ and not a variable like $x$ and $y$?
– Carlo Beenakker
9 hours ago
2
2
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
Yes, it is a fixed parameter of the system, like $k$. We should somehow use the necessary condition from the statement.
– Tina
9 hours ago
add a comment |Â
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