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Which monomials in $p$ and $q$ can be a probability in a finite coin experiment?
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The question generate event with $6.75space p^2q$, $20space p^3q^2$, $3.9space pq$? prompted me to think more generally about which expressions of this form can be a probability in a finite coin experiment. Since the answer turns out to be rather nice, I thought I'd post it here as an answer to the general question.
We have a coin that shows heads with probability $p$ and tails with probability $q:=1-p$, and the question is whether for given $muinmathbb R^+$ and $i,jinmathbb N$ (not both zero) there is an event in an experiment with a fixed finite number of tosses of the coin whose probability is $mu p^iq^j$ for all $pin[0,1]$.
probability binomial-coefficients binomial-distribution
asked Jul 16 at 21:33
joriki
164k10180328
add a comment |Â
up vote
1
down vote
favorite
The question generate event with $6.75space p^2q$, $20space p^3q^2$, $3.9space pq$? prompted me to think more generally about which expressions of this form can be a probability in a finite coin experiment. Since the answer turns out to be rather nice, I thought I'd post it here as an answer to the general question.
We have a coin that shows heads with probability $p$ and tails with probability $q:=1-p$, and the question is whether for given $muinmathbb R^+$ and $i,jinmathbb N$ (not both zero) there is an event in an experiment with a fixed finite number of tosses of the coin whose probability is $mu p^iq^j$ for all $pin[0,1]$.
probability binomial-coefficients binomial-distribution
asked Jul 16 at 21:33
joriki
164k10180328
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question generate event with $6.75space p^2q$, $20space p^3q^2$, $3.9space pq$? prompted me to think more generally about which expressions of this form can be a probability in a finite coin experiment. Since the answer turns out to be rather nice, I thought I'd post it here as an answer to the general question.
We have a coin that shows heads with probability $p$ and tails with probability $q:=1-p$, and the question is whether for given $muinmathbb R^+$ and $i,jinmathbb N$ (not both zero) there is an event in an experiment with a fixed finite number of tosses of the coin whose probability is $mu p^iq^j$ for all $pin[0,1]$.
probability binomial-coefficients binomial-distribution
asked Jul 16 at 21:33
joriki
164k10180328
The question generate event with $6.75space p^2q$, $20space p^3q^2$, $3.9space pq$? prompted me to think more generally about which expressions of this form can be a probability in a finite coin experiment. Since the answer turns out to be rather nice, I thought I'd post it here as an answer to the general question.
We have a coin that shows heads with probability $p$ and tails with probability $q:=1-p$, and the question is whether for given $muinmathbb R^+$ and $i,jinmathbb N$ (not both zero) there is an event in an experiment with a fixed finite number of tosses of the coin whose probability is $mu p^iq^j$ for all $pin[0,1]$.
probability binomial-coefficients binomial-distribution
asked Jul 16 at 21:33
joriki
164k10180328
asked Jul 16 at 21:33
joriki
164k10180328
asked Jul 16 at 21:33
joriki
164k10180328
asked Jul 16 at 21:33
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The monomial $mu p^iq^j$ is a probability in a finite coin experiment if and only if $mu$ is an integer and $mu p^iq^jlt1$ for all $pin[0,1]$.
First, $mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $mu p^iq^j=mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $mu$ for $p^i$, so $mu$ must be integer for the two to be equal.
The rest of the “only if†direction is also readily proved. Since $mu p^iq^j$ is a probability for all $p$, it must be $le1$ for all $p$. If there were $p$ such that $mu p^iq^j=1$, the event would have to include all elementary events, which implies $mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.
Now, to prove the “if†direction, consider an experiment with $i+j+l$ tosses. If we can select $mubinom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,ldots,l$, then the probability of the event composed of all these elementary events is
$$
sum_k=0^lmubinom lkp^i+kq^j+l-k=mu p^iq^j(p+q)^l=mu p^iq^j;.
$$
There are $binomi+j+li+k$ such elementary events, so we can choose $mubinom lk$ of them if
$$
mubinom lklebinomi+j+li+k
$$
for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes
$$
mu(k+i)cdots(k+1)(l-k+j)cdots(l-k+1)le(l+i+j)cdots(l+1);,
$$
and then dividing through by $l^i+j$ and denoting $frac klin[0,1]$ by $alpha$ leads to
$$
muleft(alpha+frac ilright)cdotsleft(alpha+frac1lright)left(1-alpha+frac jlright)cdotsleft(1-alpha+frac 1lright)leleft(1+fraci+jlright)cdotsleft(1+frac1lright)
$$
and thus to
$$
mualpha^i(1-alpha)^j+Oleft(frac(i+j)^2lright)le1;.
$$
Thus, we can satisfy this inequality for sufficiently large $l$ if $mualpha^i(1-alpha)^jlt1$ for all $alphain[0,1]$.
answered Jul 16 at 21:33
joriki
164k10180328
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The monomial $mu p^iq^j$ is a probability in a finite coin experiment if and only if $mu$ is an integer and $mu p^iq^jlt1$ for all $pin[0,1]$.
First, $mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $mu p^iq^j=mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $mu$ for $p^i$, so $mu$ must be integer for the two to be equal.
The rest of the “only if†direction is also readily proved. Since $mu p^iq^j$ is a probability for all $p$, it must be $le1$ for all $p$. If there were $p$ such that $mu p^iq^j=1$, the event would have to include all elementary events, which implies $mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.
Now, to prove the “if†direction, consider an experiment with $i+j+l$ tosses. If we can select $mubinom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,ldots,l$, then the probability of the event composed of all these elementary events is
$$
sum_k=0^lmubinom lkp^i+kq^j+l-k=mu p^iq^j(p+q)^l=mu p^iq^j;.
$$
There are $binomi+j+li+k$ such elementary events, so we can choose $mubinom lk$ of them if
$$
mubinom lklebinomi+j+li+k
$$
for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes
$$
mu(k+i)cdots(k+1)(l-k+j)cdots(l-k+1)le(l+i+j)cdots(l+1);,
$$
and then dividing through by $l^i+j$ and denoting $frac klin[0,1]$ by $alpha$ leads to
$$
muleft(alpha+frac ilright)cdotsleft(alpha+frac1lright)left(1-alpha+frac jlright)cdotsleft(1-alpha+frac 1lright)leleft(1+fraci+jlright)cdotsleft(1+frac1lright)
$$
and thus to
$$
mualpha^i(1-alpha)^j+Oleft(frac(i+j)^2lright)le1;.
$$
Thus, we can satisfy this inequality for sufficiently large $l$ if $mualpha^i(1-alpha)^jlt1$ for all $alphain[0,1]$.
answered Jul 16 at 21:33
joriki
164k10180328
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
add a comment |Â
up vote
2
down vote
accepted
The monomial $mu p^iq^j$ is a probability in a finite coin experiment if and only if $mu$ is an integer and $mu p^iq^jlt1$ for all $pin[0,1]$.
First, $mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $mu p^iq^j=mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $mu$ for $p^i$, so $mu$ must be integer for the two to be equal.
The rest of the “only if†direction is also readily proved. Since $mu p^iq^j$ is a probability for all $p$, it must be $le1$ for all $p$. If there were $p$ such that $mu p^iq^j=1$, the event would have to include all elementary events, which implies $mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.
Now, to prove the “if†direction, consider an experiment with $i+j+l$ tosses. If we can select $mubinom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,ldots,l$, then the probability of the event composed of all these elementary events is
$$
sum_k=0^lmubinom lkp^i+kq^j+l-k=mu p^iq^j(p+q)^l=mu p^iq^j;.
$$
There are $binomi+j+li+k$ such elementary events, so we can choose $mubinom lk$ of them if
$$
mubinom lklebinomi+j+li+k
$$
for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes
$$
mu(k+i)cdots(k+1)(l-k+j)cdots(l-k+1)le(l+i+j)cdots(l+1);,
$$
and then dividing through by $l^i+j$ and denoting $frac klin[0,1]$ by $alpha$ leads to
$$
muleft(alpha+frac ilright)cdotsleft(alpha+frac1lright)left(1-alpha+frac jlright)cdotsleft(1-alpha+frac 1lright)leleft(1+fraci+jlright)cdotsleft(1+frac1lright)
$$
and thus to
$$
mualpha^i(1-alpha)^j+Oleft(frac(i+j)^2lright)le1;.
$$
Thus, we can satisfy this inequality for sufficiently large $l$ if $mualpha^i(1-alpha)^jlt1$ for all $alphain[0,1]$.
answered Jul 16 at 21:33
joriki
164k10180328
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The monomial $mu p^iq^j$ is a probability in a finite coin experiment if and only if $mu$ is an integer and $mu p^iq^jlt1$ for all $pin[0,1]$.
First, $mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $mu p^iq^j=mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $mu$ for $p^i$, so $mu$ must be integer for the two to be equal.
The rest of the “only if†direction is also readily proved. Since $mu p^iq^j$ is a probability for all $p$, it must be $le1$ for all $p$. If there were $p$ such that $mu p^iq^j=1$, the event would have to include all elementary events, which implies $mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.
Now, to prove the “if†direction, consider an experiment with $i+j+l$ tosses. If we can select $mubinom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,ldots,l$, then the probability of the event composed of all these elementary events is
$$
sum_k=0^lmubinom lkp^i+kq^j+l-k=mu p^iq^j(p+q)^l=mu p^iq^j;.
$$
There are $binomi+j+li+k$ such elementary events, so we can choose $mubinom lk$ of them if
$$
mubinom lklebinomi+j+li+k
$$
for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes
$$
mu(k+i)cdots(k+1)(l-k+j)cdots(l-k+1)le(l+i+j)cdots(l+1);,
$$
and then dividing through by $l^i+j$ and denoting $frac klin[0,1]$ by $alpha$ leads to
$$
muleft(alpha+frac ilright)cdotsleft(alpha+frac1lright)left(1-alpha+frac jlright)cdotsleft(1-alpha+frac 1lright)leleft(1+fraci+jlright)cdotsleft(1+frac1lright)
$$
and thus to
$$
mualpha^i(1-alpha)^j+Oleft(frac(i+j)^2lright)le1;.
$$
Thus, we can satisfy this inequality for sufficiently large $l$ if $mualpha^i(1-alpha)^jlt1$ for all $alphain[0,1]$.
answered Jul 16 at 21:33
joriki
164k10180328
The monomial $mu p^iq^j$ is a probability in a finite coin experiment if and only if $mu$ is an integer and $mu p^iq^jlt1$ for all $pin[0,1]$.
First, $mu$ needs to be an integer: The probability of any event in an experiment with a fixed finite number of tosses is a polynomial in $p$ with integer coefficients; and $mu p^iq^j=mu p^i(1-p)^j$ is a polynomial in $p$ with coefficient $mu$ for $p^i$, so $mu$ must be integer for the two to be equal.
The rest of the “only if†direction is also readily proved. Since $mu p^iq^j$ is a probability for all $p$, it must be $le1$ for all $p$. If there were $p$ such that $mu p^iq^j=1$, the event would have to include all elementary events, which implies $mu p^iq^j=1$ for all $p$, which only occurs in the trivial case $i=j=0$ that was excluded in the question.
Now, to prove the “if†direction, consider an experiment with $i+j+l$ tosses. If we can select $mubinom lk$ elementary events with $i+k$ heads and $j+l-k$ tails for all $k=0,ldots,l$, then the probability of the event composed of all these elementary events is
$$
sum_k=0^lmubinom lkp^i+kq^j+l-k=mu p^iq^j(p+q)^l=mu p^iq^j;.
$$
There are $binomi+j+li+k$ such elementary events, so we can choose $mubinom lk$ of them if
$$
mubinom lklebinomi+j+li+k
$$
for all $k$. If we cancel factors in the three pairs of corresponding factorials, this becomes
$$
mu(k+i)cdots(k+1)(l-k+j)cdots(l-k+1)le(l+i+j)cdots(l+1);,
$$
and then dividing through by $l^i+j$ and denoting $frac klin[0,1]$ by $alpha$ leads to
$$
muleft(alpha+frac ilright)cdotsleft(alpha+frac1lright)left(1-alpha+frac jlright)cdotsleft(1-alpha+frac 1lright)leleft(1+fraci+jlright)cdotsleft(1+frac1lright)
$$
and thus to
$$
mualpha^i(1-alpha)^j+Oleft(frac(i+j)^2lright)le1;.
$$
Thus, we can satisfy this inequality for sufficiently large $l$ if $mualpha^i(1-alpha)^jlt1$ for all $alphain[0,1]$.
answered Jul 16 at 21:33
joriki
164k10180328
answered Jul 16 at 21:33
joriki
164k10180328
answered Jul 16 at 21:33
joriki
164k10180328
answered Jul 16 at 21:33
joriki
164k10180328
164k10180328
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
add a comment |Â
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
but, if you take an event from a subset of the (finite) probability space, can't $mu$ be rational ? and even greater than one , keeping safe that $mu p^i q^j <1$ ?
– G Cab
Jul 16 at 21:49
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
@GCab: $mu$ can certainly be greater than $1$; in the original question that I linked to, it was $20$. But I don't understand your question about $mu$ being rational. Is there an error in my proof that it can't? Perhaps I don't understand what you mean by "taking an event from a subset".
– joriki
Jul 16 at 21:54
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
My doubt was about introducing order. But I got your point now, in a different way: let's say that each of all the possible binary strings of length $n$ has a probability that corresponds to the product of $(p_1p_2cdots p_m,;q_1q_2cdots q_n-m)$ in whichever order. So $p_iq_j$ is the .."granularity" of any event you can define in them, and so $mu$ must be an integer. (+1)
– G Cab
Jul 16 at 23:50
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
@GCab: I don't think the argument works in quite that way. As the proof here and already the example given in the linked question illustrate, you can make use of $p+q=1$ to combine monomials into lower ones; and you can also use it to artificially increase the degree of a polynomial in $p$ and $q$; so I don't see how working with $p$ and $q$ allows for such a "granularity" argument; I think you really do need to switch to a representation in terms of $p$ alone, as I did in the proof, in order to make that argument go through.
– joriki
Jul 17 at 3:56
add a comment |Â
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