Mixing
Use of a Joint Moment Generation Function - Do I have this right?
Clash Royale CLAN TAG#URR8PPP
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Below is a problem I did. I am confident that I got part a right. I am not
confident that I got part b right. In particular, I am thinking that my use of
the partial derivative symbol may not be right. I am hoping that somebody can
check my work.
Thanks
Bob
Problem:
Let $(X,Y)$ be a continues bivariate r.v. with joint pdf
begineqnarray*
f_XY(x,y) &=& begincases
e^-(x+y) & x > 0 , y > 0 \
0 & textotherwise \
endcases \
endeqnarray*
(a) Find the joint moment generating function of $X$ and $Y$.
(b) Find the joint moments $m_10$, $m_01$ and $m_11$.
Answer: (a)
begineqnarray*
M_XY &=& E(e^t_1 X + t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + t_2 y)e^-(x+y)
, dy , dx \
M_XY &=&
int_0^infty int_0^infty e^t_1x - x + t_2y -y , dy , dx \
M_XY &=&int_0^infty frace^t_1x - x + t_2y -yt_2 - 1 ,
Big|_y = 0^y = infty dx \
M_XY &=&int_0^infty 0 - frace^t_1x - xt_2 - 1 , dx \
M_XY &=&int_0^infty frace^t_1x - x1 - t_2 , dx \
M_XY &=& frace^t_1x - x(t_1+ 1)(1 - t_2) Big|_0^infty \
M_XY &=& 0 - frac1(t_1 - 1)(1 - t_2) \
M_XY &=& frac1 (t_1 - 1)(t_2 - 1) \
M_XY &=& frac1(1 - t_1)(1 - t_2) \
endeqnarray*
Part (b)
begineqnarray*
M_XY &=& (t_1 - 1 )^-1 (t_2 - 1)^-1\
m_10 &=& fracpartialpartial t_1 M_XY(0,0) \
fracpartialpartial t_1 M_XY &=& -(t_1-1)^-2(t_2-1)^-1 \
fracpartialpartial t_1 M_XY(0,0) &=& -(0-1)^-2(0-1)^-1 = -(1)(-1) \
m_10 &=& 1 \
m_01 &=& fracpartialpartial t_2 M_XY(0,0) \
fracpartialpartial t_2 M_XY &=& -(t_1-1)^-1(t_2-1)^-2 \
fracpartialpartial t_2 M_XY(0,0) &=& -(0 - 1)^-1(0-1)^-2 = -(-1)(1) \
fracpartialpartial t_2 M_XY(0,0) &=& 1 \
m_01 &=& 1 \
m_11 &=& fracpartial^2partial t_1 partial t_2 M_XY(0,0) \
fracpartial^2partial t_1 partial t_2 M_XY &=&
(t_1-1)^-2 (t_2 - 1)^-2 \
fracpartial^2partial t_1 partial t_2 M_XY(0,0 ) &=&
(0-1)^-2 (0 - 1)^-2 = 1(1) = 1 \
m_11 &=& 1 \
endeqnarray*
probability probability-distributions
asked Aug 3 at 18:51
Bob
717410
add a comment |Â
up vote
0
down vote
favorite
Below is a problem I did. I am confident that I got part a right. I am not
confident that I got part b right. In particular, I am thinking that my use of
the partial derivative symbol may not be right. I am hoping that somebody can
check my work.
Thanks
Bob
Problem:
Let $(X,Y)$ be a continues bivariate r.v. with joint pdf
begineqnarray*
f_XY(x,y) &=& begincases
e^-(x+y) & x > 0 , y > 0 \
0 & textotherwise \
endcases \
endeqnarray*
(a) Find the joint moment generating function of $X$ and $Y$.
(b) Find the joint moments $m_10$, $m_01$ and $m_11$.
Answer: (a)
begineqnarray*
M_XY &=& E(e^t_1 X + t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + t_2 y)e^-(x+y)
, dy , dx \
M_XY &=&
int_0^infty int_0^infty e^t_1x - x + t_2y -y , dy , dx \
M_XY &=&int_0^infty frace^t_1x - x + t_2y -yt_2 - 1 ,
Big|_y = 0^y = infty dx \
M_XY &=&int_0^infty 0 - frace^t_1x - xt_2 - 1 , dx \
M_XY &=&int_0^infty frace^t_1x - x1 - t_2 , dx \
M_XY &=& frace^t_1x - x(t_1+ 1)(1 - t_2) Big|_0^infty \
M_XY &=& 0 - frac1(t_1 - 1)(1 - t_2) \
M_XY &=& frac1 (t_1 - 1)(t_2 - 1) \
M_XY &=& frac1(1 - t_1)(1 - t_2) \
endeqnarray*
Part (b)
begineqnarray*
M_XY &=& (t_1 - 1 )^-1 (t_2 - 1)^-1\
m_10 &=& fracpartialpartial t_1 M_XY(0,0) \
fracpartialpartial t_1 M_XY &=& -(t_1-1)^-2(t_2-1)^-1 \
fracpartialpartial t_1 M_XY(0,0) &=& -(0-1)^-2(0-1)^-1 = -(1)(-1) \
m_10 &=& 1 \
m_01 &=& fracpartialpartial t_2 M_XY(0,0) \
fracpartialpartial t_2 M_XY &=& -(t_1-1)^-1(t_2-1)^-2 \
fracpartialpartial t_2 M_XY(0,0) &=& -(0 - 1)^-1(0-1)^-2 = -(-1)(1) \
fracpartialpartial t_2 M_XY(0,0) &=& 1 \
m_01 &=& 1 \
m_11 &=& fracpartial^2partial t_1 partial t_2 M_XY(0,0) \
fracpartial^2partial t_1 partial t_2 M_XY &=&
(t_1-1)^-2 (t_2 - 1)^-2 \
fracpartial^2partial t_1 partial t_2 M_XY(0,0 ) &=&
(0-1)^-2 (0 - 1)^-2 = 1(1) = 1 \
m_11 &=& 1 \
endeqnarray*
probability probability-distributions
asked Aug 3 at 18:51
Bob
717410
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Below is a problem I did. I am confident that I got part a right. I am not
confident that I got part b right. In particular, I am thinking that my use of
the partial derivative symbol may not be right. I am hoping that somebody can
check my work.
Thanks
Bob
Problem:
Let $(X,Y)$ be a continues bivariate r.v. with joint pdf
begineqnarray*
f_XY(x,y) &=& begincases
e^-(x+y) & x > 0 , y > 0 \
0 & textotherwise \
endcases \
endeqnarray*
(a) Find the joint moment generating function of $X$ and $Y$.
(b) Find the joint moments $m_10$, $m_01$ and $m_11$.
Answer: (a)
begineqnarray*
M_XY &=& E(e^t_1 X + t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + t_2 y)e^-(x+y)
, dy , dx \
M_XY &=&
int_0^infty int_0^infty e^t_1x - x + t_2y -y , dy , dx \
M_XY &=&int_0^infty frace^t_1x - x + t_2y -yt_2 - 1 ,
Big|_y = 0^y = infty dx \
M_XY &=&int_0^infty 0 - frace^t_1x - xt_2 - 1 , dx \
M_XY &=&int_0^infty frace^t_1x - x1 - t_2 , dx \
M_XY &=& frace^t_1x - x(t_1+ 1)(1 - t_2) Big|_0^infty \
M_XY &=& 0 - frac1(t_1 - 1)(1 - t_2) \
M_XY &=& frac1 (t_1 - 1)(t_2 - 1) \
M_XY &=& frac1(1 - t_1)(1 - t_2) \
endeqnarray*
Part (b)
begineqnarray*
M_XY &=& (t_1 - 1 )^-1 (t_2 - 1)^-1\
m_10 &=& fracpartialpartial t_1 M_XY(0,0) \
fracpartialpartial t_1 M_XY &=& -(t_1-1)^-2(t_2-1)^-1 \
fracpartialpartial t_1 M_XY(0,0) &=& -(0-1)^-2(0-1)^-1 = -(1)(-1) \
m_10 &=& 1 \
m_01 &=& fracpartialpartial t_2 M_XY(0,0) \
fracpartialpartial t_2 M_XY &=& -(t_1-1)^-1(t_2-1)^-2 \
fracpartialpartial t_2 M_XY(0,0) &=& -(0 - 1)^-1(0-1)^-2 = -(-1)(1) \
fracpartialpartial t_2 M_XY(0,0) &=& 1 \
m_01 &=& 1 \
m_11 &=& fracpartial^2partial t_1 partial t_2 M_XY(0,0) \
fracpartial^2partial t_1 partial t_2 M_XY &=&
(t_1-1)^-2 (t_2 - 1)^-2 \
fracpartial^2partial t_1 partial t_2 M_XY(0,0 ) &=&
(0-1)^-2 (0 - 1)^-2 = 1(1) = 1 \
m_11 &=& 1 \
endeqnarray*
probability probability-distributions
asked Aug 3 at 18:51
Bob
717410
Below is a problem I did. I am confident that I got part a right. I am not
confident that I got part b right. In particular, I am thinking that my use of
the partial derivative symbol may not be right. I am hoping that somebody can
check my work.
Thanks
Bob
Problem:
Let $(X,Y)$ be a continues bivariate r.v. with joint pdf
begineqnarray*
f_XY(x,y) &=& begincases
e^-(x+y) & x > 0 , y > 0 \
0 & textotherwise \
endcases \
endeqnarray*
(a) Find the joint moment generating function of $X$ and $Y$.
(b) Find the joint moments $m_10$, $m_01$ and $m_11$.
Answer: (a)
begineqnarray*
M_XY &=& E(e^t_1 X + t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + t_2 y)e^-(x+y)
, dy , dx \
M_XY &=&
int_0^infty int_0^infty e^t_1x - x + t_2y -y , dy , dx \
M_XY &=&int_0^infty frace^t_1x - x + t_2y -yt_2 - 1 ,
Big|_y = 0^y = infty dx \
M_XY &=&int_0^infty 0 - frace^t_1x - xt_2 - 1 , dx \
M_XY &=&int_0^infty frace^t_1x - x1 - t_2 , dx \
M_XY &=& frace^t_1x - x(t_1+ 1)(1 - t_2) Big|_0^infty \
M_XY &=& 0 - frac1(t_1 - 1)(1 - t_2) \
M_XY &=& frac1 (t_1 - 1)(t_2 - 1) \
M_XY &=& frac1(1 - t_1)(1 - t_2) \
endeqnarray*
Part (b)
begineqnarray*
M_XY &=& (t_1 - 1 )^-1 (t_2 - 1)^-1\
m_10 &=& fracpartialpartial t_1 M_XY(0,0) \
fracpartialpartial t_1 M_XY &=& -(t_1-1)^-2(t_2-1)^-1 \
fracpartialpartial t_1 M_XY(0,0) &=& -(0-1)^-2(0-1)^-1 = -(1)(-1) \
m_10 &=& 1 \
m_01 &=& fracpartialpartial t_2 M_XY(0,0) \
fracpartialpartial t_2 M_XY &=& -(t_1-1)^-1(t_2-1)^-2 \
fracpartialpartial t_2 M_XY(0,0) &=& -(0 - 1)^-1(0-1)^-2 = -(-1)(1) \
fracpartialpartial t_2 M_XY(0,0) &=& 1 \
m_01 &=& 1 \
m_11 &=& fracpartial^2partial t_1 partial t_2 M_XY(0,0) \
fracpartial^2partial t_1 partial t_2 M_XY &=&
(t_1-1)^-2 (t_2 - 1)^-2 \
fracpartial^2partial t_1 partial t_2 M_XY(0,0 ) &=&
(0-1)^-2 (0 - 1)^-2 = 1(1) = 1 \
m_11 &=& 1 \
endeqnarray*
probability probability-distributions
asked Aug 3 at 18:51
Bob
717410
edited Aug 3 at 19:18
asked Aug 3 at 18:51
Bob
717410
asked Aug 3 at 18:51
Bob
717410
asked Aug 3 at 18:51
Bob
717410
717410
add a comment |Â
add a comment |Â
1 Answer
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Some feedback:
I haven't looked at your work in extreme detail, but it looks mostly good.
I would prefer that you use $f_X, Y$ rather than $f_XY$ to make it clear that it is the joint density, rather than the density of the product $XY$.
I know that some probability textbooks use the notation
$$dfracpartialpartial t_1M_X,Y(0, 0)$$
but I prefer
$$left.dfracpartialpartial t_1M_X,Y(t_1, t_2)right|_(t_1, t_2) = (0, 0)$$
because this notation makes it more clear that the $(0, 0)$ needs to be plugged in after the partial derivative is calculated.
Besides the notational stuff, your work does look good. But there's a much shorter way to do this.
Notice that $$f_X, Y(x, y) = e^-xe^y = f_X(x)f_Y(y)$$
for $x, y > 0$, so by this factorization, it follows that $X$ and $Y$ must be independent exponential random variables with mean $1$.
Thus, it follows that $e^t_1X$ and $e^t_2Y$ are independent, and you may use the fact that expectation is multiplicative when random variables are independent:
$$M_X, Y(t_1, t_2) = mathbbE[e^t_1X+t_2Y]=mathbbE[e^t_1Xe^t_2Y]=mathbbE[e^t_1X]mathbbE[e^t_2Y]=dfrac1(1-t_1)(1-t_2)$$
(assuming you already know the MGF of an exponential distribution) for $t_1, t_2$ chosen appropriately (which I will leave for you to find).
It follows easily that $m_10 = m_01 = 1$, and since we've established that $X$ and $Y$ are independent,
$$m_11 = mathbbE[XY]=mathbbE[X]mathbbE[Y]=1 cdot 1 = 1text.$$
answered Aug 3 at 19:19
Clarinetist
10.2k32767
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Some feedback:
I haven't looked at your work in extreme detail, but it looks mostly good.
I would prefer that you use $f_X, Y$ rather than $f_XY$ to make it clear that it is the joint density, rather than the density of the product $XY$.
I know that some probability textbooks use the notation
$$dfracpartialpartial t_1M_X,Y(0, 0)$$
but I prefer
$$left.dfracpartialpartial t_1M_X,Y(t_1, t_2)right|_(t_1, t_2) = (0, 0)$$
because this notation makes it more clear that the $(0, 0)$ needs to be plugged in after the partial derivative is calculated.
Besides the notational stuff, your work does look good. But there's a much shorter way to do this.
Notice that $$f_X, Y(x, y) = e^-xe^y = f_X(x)f_Y(y)$$
for $x, y > 0$, so by this factorization, it follows that $X$ and $Y$ must be independent exponential random variables with mean $1$.
Thus, it follows that $e^t_1X$ and $e^t_2Y$ are independent, and you may use the fact that expectation is multiplicative when random variables are independent:
$$M_X, Y(t_1, t_2) = mathbbE[e^t_1X+t_2Y]=mathbbE[e^t_1Xe^t_2Y]=mathbbE[e^t_1X]mathbbE[e^t_2Y]=dfrac1(1-t_1)(1-t_2)$$
(assuming you already know the MGF of an exponential distribution) for $t_1, t_2$ chosen appropriately (which I will leave for you to find).
It follows easily that $m_10 = m_01 = 1$, and since we've established that $X$ and $Y$ are independent,
$$m_11 = mathbbE[XY]=mathbbE[X]mathbbE[Y]=1 cdot 1 = 1text.$$
answered Aug 3 at 19:19
Clarinetist
10.2k32767
add a comment |Â
up vote
0
down vote
accepted
Some feedback:
I haven't looked at your work in extreme detail, but it looks mostly good.
I would prefer that you use $f_X, Y$ rather than $f_XY$ to make it clear that it is the joint density, rather than the density of the product $XY$.
I know that some probability textbooks use the notation
$$dfracpartialpartial t_1M_X,Y(0, 0)$$
but I prefer
$$left.dfracpartialpartial t_1M_X,Y(t_1, t_2)right|_(t_1, t_2) = (0, 0)$$
because this notation makes it more clear that the $(0, 0)$ needs to be plugged in after the partial derivative is calculated.
Besides the notational stuff, your work does look good. But there's a much shorter way to do this.
Notice that $$f_X, Y(x, y) = e^-xe^y = f_X(x)f_Y(y)$$
for $x, y > 0$, so by this factorization, it follows that $X$ and $Y$ must be independent exponential random variables with mean $1$.
Thus, it follows that $e^t_1X$ and $e^t_2Y$ are independent, and you may use the fact that expectation is multiplicative when random variables are independent:
$$M_X, Y(t_1, t_2) = mathbbE[e^t_1X+t_2Y]=mathbbE[e^t_1Xe^t_2Y]=mathbbE[e^t_1X]mathbbE[e^t_2Y]=dfrac1(1-t_1)(1-t_2)$$
(assuming you already know the MGF of an exponential distribution) for $t_1, t_2$ chosen appropriately (which I will leave for you to find).
It follows easily that $m_10 = m_01 = 1$, and since we've established that $X$ and $Y$ are independent,
$$m_11 = mathbbE[XY]=mathbbE[X]mathbbE[Y]=1 cdot 1 = 1text.$$
answered Aug 3 at 19:19
Clarinetist
10.2k32767
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Some feedback:
I haven't looked at your work in extreme detail, but it looks mostly good.
I would prefer that you use $f_X, Y$ rather than $f_XY$ to make it clear that it is the joint density, rather than the density of the product $XY$.
I know that some probability textbooks use the notation
$$dfracpartialpartial t_1M_X,Y(0, 0)$$
but I prefer
$$left.dfracpartialpartial t_1M_X,Y(t_1, t_2)right|_(t_1, t_2) = (0, 0)$$
because this notation makes it more clear that the $(0, 0)$ needs to be plugged in after the partial derivative is calculated.
Besides the notational stuff, your work does look good. But there's a much shorter way to do this.
Notice that $$f_X, Y(x, y) = e^-xe^y = f_X(x)f_Y(y)$$
for $x, y > 0$, so by this factorization, it follows that $X$ and $Y$ must be independent exponential random variables with mean $1$.
Thus, it follows that $e^t_1X$ and $e^t_2Y$ are independent, and you may use the fact that expectation is multiplicative when random variables are independent:
$$M_X, Y(t_1, t_2) = mathbbE[e^t_1X+t_2Y]=mathbbE[e^t_1Xe^t_2Y]=mathbbE[e^t_1X]mathbbE[e^t_2Y]=dfrac1(1-t_1)(1-t_2)$$
(assuming you already know the MGF of an exponential distribution) for $t_1, t_2$ chosen appropriately (which I will leave for you to find).
It follows easily that $m_10 = m_01 = 1$, and since we've established that $X$ and $Y$ are independent,
$$m_11 = mathbbE[XY]=mathbbE[X]mathbbE[Y]=1 cdot 1 = 1text.$$
answered Aug 3 at 19:19
Clarinetist
10.2k32767
Some feedback:
I haven't looked at your work in extreme detail, but it looks mostly good.
I would prefer that you use $f_X, Y$ rather than $f_XY$ to make it clear that it is the joint density, rather than the density of the product $XY$.
I know that some probability textbooks use the notation
$$dfracpartialpartial t_1M_X,Y(0, 0)$$
but I prefer
$$left.dfracpartialpartial t_1M_X,Y(t_1, t_2)right|_(t_1, t_2) = (0, 0)$$
because this notation makes it more clear that the $(0, 0)$ needs to be plugged in after the partial derivative is calculated.
Besides the notational stuff, your work does look good. But there's a much shorter way to do this.
Notice that $$f_X, Y(x, y) = e^-xe^y = f_X(x)f_Y(y)$$
for $x, y > 0$, so by this factorization, it follows that $X$ and $Y$ must be independent exponential random variables with mean $1$.
Thus, it follows that $e^t_1X$ and $e^t_2Y$ are independent, and you may use the fact that expectation is multiplicative when random variables are independent:
$$M_X, Y(t_1, t_2) = mathbbE[e^t_1X+t_2Y]=mathbbE[e^t_1Xe^t_2Y]=mathbbE[e^t_1X]mathbbE[e^t_2Y]=dfrac1(1-t_1)(1-t_2)$$
(assuming you already know the MGF of an exponential distribution) for $t_1, t_2$ chosen appropriately (which I will leave for you to find).
It follows easily that $m_10 = m_01 = 1$, and since we've established that $X$ and $Y$ are independent,
$$m_11 = mathbbE[XY]=mathbbE[X]mathbbE[Y]=1 cdot 1 = 1text.$$
answered Aug 3 at 19:19
Clarinetist
10.2k32767
answered Aug 3 at 19:19
Clarinetist
10.2k32767
answered Aug 3 at 19:19
Clarinetist
10.2k32767
answered Aug 3 at 19:19
Clarinetist
10.2k32767
10.2k32767
add a comment |Â
add a comment |Â
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