Mixing
Finding a Joint Moment Generating Function
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Please consider the following problem and my solution to it:
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*
Find the joint moment generating function of $X$ and $Y$.
Answer:
begineqnarray*
M_XY &=& E(e^t_1 X + e^t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + e^t_2 y)(e^-(x+y)) , dy , dx \
M_XY &=&
int_0^infty int_0^infty
e^t_1x - x - y + e^t_2y - x - y , dy , dx \
M_XY &=&
int_0^infty
-e^t_1x - x - y - frac e^t_2y -x - y t_2-1 Big|_y = 0^y = infty , dx \
M_XY &=& int_0^infty
( -0 + 0) - ( -e^t_1x - x - frace^ -x t_2-1 ) , dx \
M_XY &=& int_0^infty e^(t_1-1)x + frac e^-xt_2-1 , dx \
M_XY &=&
frace^(t_1-1)xt_1-1 - frac e^ -x t_2-1
Big|_x = 0^x = infty \
M_XY &=& ( 0 + 0 ) - ( frac1t_1 - 1 - frac1t_2 - 1) \
M_XY &=& frac1t_2 - 1 - frac1t_1 - 1 \
M_XY &=& frac t_1 - 1 - ( t_2 - 1 ) (t_1 - 1)(t_2 - 1) \
M_XY &=& frac t_1 - t_2 (t_1 - 1)(t_2 - 1) \
endeqnarray*
However, the book gets:
begineqnarray*
M_XY &=& frac1 (1 - t_1)(1 - t_2) \
endeqnarray*
Please note that my answer could be rewritten as:
begineqnarray*
M_XY &=& fract_1-t_2 (1 - t_1)(1 - t_2) \
endeqnarray*
What did I do wrong?
Thanks,
Bob
probability probability-distributions moment-generating-functions
edited Aug 1 at 22:27
Clement C.
46.9k33682
asked Aug 1 at 22:06
Bob
720411
add a comment |Â
up vote
1
down vote
favorite
Please consider the following problem and my solution to it:
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*
Find the joint moment generating function of $X$ and $Y$.
Answer:
begineqnarray*
M_XY &=& E(e^t_1 X + e^t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + e^t_2 y)(e^-(x+y)) , dy , dx \
M_XY &=&
int_0^infty int_0^infty
e^t_1x - x - y + e^t_2y - x - y , dy , dx \
M_XY &=&
int_0^infty
-e^t_1x - x - y - frac e^t_2y -x - y t_2-1 Big|_y = 0^y = infty , dx \
M_XY &=& int_0^infty
( -0 + 0) - ( -e^t_1x - x - frace^ -x t_2-1 ) , dx \
M_XY &=& int_0^infty e^(t_1-1)x + frac e^-xt_2-1 , dx \
M_XY &=&
frace^(t_1-1)xt_1-1 - frac e^ -x t_2-1
Big|_x = 0^x = infty \
M_XY &=& ( 0 + 0 ) - ( frac1t_1 - 1 - frac1t_2 - 1) \
M_XY &=& frac1t_2 - 1 - frac1t_1 - 1 \
M_XY &=& frac t_1 - 1 - ( t_2 - 1 ) (t_1 - 1)(t_2 - 1) \
M_XY &=& frac t_1 - t_2 (t_1 - 1)(t_2 - 1) \
endeqnarray*
However, the book gets:
begineqnarray*
M_XY &=& frac1 (1 - t_1)(1 - t_2) \
endeqnarray*
Please note that my answer could be rewritten as:
begineqnarray*
M_XY &=& fract_1-t_2 (1 - t_1)(1 - t_2) \
endeqnarray*
What did I do wrong?
Thanks,
Bob
probability probability-distributions moment-generating-functions
edited Aug 1 at 22:27
Clement C.
46.9k33682
asked Aug 1 at 22:06
Bob
720411
1
As $X$ and $Y$ are independent Exponential$(1)$ variables, the joint MGF is the product of the individual MGFs.
– StubbornAtom
Aug 2 at 15:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Please consider the following problem and my solution to it:
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*
Find the joint moment generating function of $X$ and $Y$.
Answer:
begineqnarray*
M_XY &=& E(e^t_1 X + e^t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + e^t_2 y)(e^-(x+y)) , dy , dx \
M_XY &=&
int_0^infty int_0^infty
e^t_1x - x - y + e^t_2y - x - y , dy , dx \
M_XY &=&
int_0^infty
-e^t_1x - x - y - frac e^t_2y -x - y t_2-1 Big|_y = 0^y = infty , dx \
M_XY &=& int_0^infty
( -0 + 0) - ( -e^t_1x - x - frace^ -x t_2-1 ) , dx \
M_XY &=& int_0^infty e^(t_1-1)x + frac e^-xt_2-1 , dx \
M_XY &=&
frace^(t_1-1)xt_1-1 - frac e^ -x t_2-1
Big|_x = 0^x = infty \
M_XY &=& ( 0 + 0 ) - ( frac1t_1 - 1 - frac1t_2 - 1) \
M_XY &=& frac1t_2 - 1 - frac1t_1 - 1 \
M_XY &=& frac t_1 - 1 - ( t_2 - 1 ) (t_1 - 1)(t_2 - 1) \
M_XY &=& frac t_1 - t_2 (t_1 - 1)(t_2 - 1) \
endeqnarray*
However, the book gets:
begineqnarray*
M_XY &=& frac1 (1 - t_1)(1 - t_2) \
endeqnarray*
Please note that my answer could be rewritten as:
begineqnarray*
M_XY &=& fract_1-t_2 (1 - t_1)(1 - t_2) \
endeqnarray*
What did I do wrong?
Thanks,
Bob
probability probability-distributions moment-generating-functions
edited Aug 1 at 22:27
Clement C.
46.9k33682
asked Aug 1 at 22:06
Bob
720411
Please consider the following problem and my solution to it:
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*
Find the joint moment generating function of $X$ and $Y$.
Answer:
begineqnarray*
M_XY &=& E(e^t_1 X + e^t_2 Y) \
M_XY &=&
int_0^infty int_0^infty (e^t_1 x + e^t_2 y)(e^-(x+y)) , dy , dx \
M_XY &=&
int_0^infty int_0^infty
e^t_1x - x - y + e^t_2y - x - y , dy , dx \
M_XY &=&
int_0^infty
-e^t_1x - x - y - frac e^t_2y -x - y t_2-1 Big|_y = 0^y = infty , dx \
M_XY &=& int_0^infty
( -0 + 0) - ( -e^t_1x - x - frace^ -x t_2-1 ) , dx \
M_XY &=& int_0^infty e^(t_1-1)x + frac e^-xt_2-1 , dx \
M_XY &=&
frace^(t_1-1)xt_1-1 - frac e^ -x t_2-1
Big|_x = 0^x = infty \
M_XY &=& ( 0 + 0 ) - ( frac1t_1 - 1 - frac1t_2 - 1) \
M_XY &=& frac1t_2 - 1 - frac1t_1 - 1 \
M_XY &=& frac t_1 - 1 - ( t_2 - 1 ) (t_1 - 1)(t_2 - 1) \
M_XY &=& frac t_1 - t_2 (t_1 - 1)(t_2 - 1) \
endeqnarray*
However, the book gets:
begineqnarray*
M_XY &=& frac1 (1 - t_1)(1 - t_2) \
endeqnarray*
Please note that my answer could be rewritten as:
begineqnarray*
M_XY &=& fract_1-t_2 (1 - t_1)(1 - t_2) \
endeqnarray*
What did I do wrong?
Thanks,
Bob
probability probability-distributions moment-generating-functions
edited Aug 1 at 22:27
Clement C.
46.9k33682
asked Aug 1 at 22:06
Bob
720411
edited Aug 1 at 22:27
Clement C.
46.9k33682
edited Aug 1 at 22:27
Clement C.
46.9k33682
edited Aug 1 at 22:27
Clement C.
46.9k33682
46.9k33682
asked Aug 1 at 22:06
Bob
720411
asked Aug 1 at 22:06
Bob
720411
asked Aug 1 at 22:06
Bob
720411
720411
1
As $X$ and $Y$ are independent Exponential$(1)$ variables, the joint MGF is the product of the individual MGFs.
– StubbornAtom
Aug 2 at 15:50
add a comment |Â
1
As $X$ and $Y$ are independent Exponential$(1)$ variables, the joint MGF is the product of the individual MGFs.
– StubbornAtom
Aug 2 at 15:50
1
1
As $X$ and $Y$ are independent Exponential$(1)$ variables, the joint MGF is the product of the individual MGFs.
– StubbornAtom
Aug 2 at 15:50
As $X$ and $Y$ are independent Exponential$(1)$ variables, the joint MGF is the product of the individual MGFs.
– StubbornAtom
Aug 2 at 15:50
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
What is wrong is your expression for the MGF. It should be
$$
M_XY(t_1,t_2) = mathbbE[e^biglangle beginpmatrix t_1\t_2endpmatrix,beginpmatrix X\Yendpmatrix bigrangle]
= mathbbE[e^t_1X+t_2Y] tag$dagger$
$$
not $mathbbE[e^t_1X+e^t_2Y]$ (which, by linearity, would always be equal to $mathbbE[e^t_1X]+mathbbE[e^t_2Y]$—this should strike you as strange).
Using the expression from $(dagger)$, you 'll get, for $t_1,t_2<1$,
$$beginalign
M_XY(t_1,t_2) &= int_0^infty int_0^infty dxdy e^t_1x+t_2ye^-(x+y)
= int_0^infty int_0^infty dxdy e^(t_1-1)xe^(t_2-1)y\
&= int_0^infty dx e^(t_1-1)x int_0^infty dye^(t_2-1)y
= frac11-t_1 cdot frac11-t_2
endalign$$
as expected.
answered Aug 1 at 22:19
Clement C.
46.9k33682
add a comment |Â
up vote
0
down vote
You seek $$beginalignmathsf M_X,Y(s,t) &= mathsf E(mathrm e^sX+tY) \ &= mathsf E(mathrm e^sX~mathrm e^tY)\ &=mathsf E(mathrm e^sX)~mathsf E(mathrm e^tY) & star\ & =mathsf M_X(s)~mathsf M_Y(t)endalign$$
$star~$ Since $f_X,Y(x,y)=mathrm e^-xmathrm e^-y mathbf 1_xgeq 0mathbf 1_ygeq 0$ indicates that the random variables are independent, and infact something. Â Show this to be so, and thus use the MGF for that distribution to find the joint MGF.
answered Aug 1 at 23:36
Graham Kemp
80k43275
2
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
What is wrong is your expression for the MGF. It should be
$$
M_XY(t_1,t_2) = mathbbE[e^biglangle beginpmatrix t_1\t_2endpmatrix,beginpmatrix X\Yendpmatrix bigrangle]
= mathbbE[e^t_1X+t_2Y] tag$dagger$
$$
not $mathbbE[e^t_1X+e^t_2Y]$ (which, by linearity, would always be equal to $mathbbE[e^t_1X]+mathbbE[e^t_2Y]$—this should strike you as strange).
Using the expression from $(dagger)$, you 'll get, for $t_1,t_2<1$,
$$beginalign
M_XY(t_1,t_2) &= int_0^infty int_0^infty dxdy e^t_1x+t_2ye^-(x+y)
= int_0^infty int_0^infty dxdy e^(t_1-1)xe^(t_2-1)y\
&= int_0^infty dx e^(t_1-1)x int_0^infty dye^(t_2-1)y
= frac11-t_1 cdot frac11-t_2
endalign$$
as expected.
answered Aug 1 at 22:19
Clement C.
46.9k33682
add a comment |Â
up vote
2
down vote
accepted
What is wrong is your expression for the MGF. It should be
$$
M_XY(t_1,t_2) = mathbbE[e^biglangle beginpmatrix t_1\t_2endpmatrix,beginpmatrix X\Yendpmatrix bigrangle]
= mathbbE[e^t_1X+t_2Y] tag$dagger$
$$
not $mathbbE[e^t_1X+e^t_2Y]$ (which, by linearity, would always be equal to $mathbbE[e^t_1X]+mathbbE[e^t_2Y]$—this should strike you as strange).
Using the expression from $(dagger)$, you 'll get, for $t_1,t_2<1$,
$$beginalign
M_XY(t_1,t_2) &= int_0^infty int_0^infty dxdy e^t_1x+t_2ye^-(x+y)
= int_0^infty int_0^infty dxdy e^(t_1-1)xe^(t_2-1)y\
&= int_0^infty dx e^(t_1-1)x int_0^infty dye^(t_2-1)y
= frac11-t_1 cdot frac11-t_2
endalign$$
as expected.
answered Aug 1 at 22:19
Clement C.
46.9k33682
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
What is wrong is your expression for the MGF. It should be
$$
M_XY(t_1,t_2) = mathbbE[e^biglangle beginpmatrix t_1\t_2endpmatrix,beginpmatrix X\Yendpmatrix bigrangle]
= mathbbE[e^t_1X+t_2Y] tag$dagger$
$$
not $mathbbE[e^t_1X+e^t_2Y]$ (which, by linearity, would always be equal to $mathbbE[e^t_1X]+mathbbE[e^t_2Y]$—this should strike you as strange).
Using the expression from $(dagger)$, you 'll get, for $t_1,t_2<1$,
$$beginalign
M_XY(t_1,t_2) &= int_0^infty int_0^infty dxdy e^t_1x+t_2ye^-(x+y)
= int_0^infty int_0^infty dxdy e^(t_1-1)xe^(t_2-1)y\
&= int_0^infty dx e^(t_1-1)x int_0^infty dye^(t_2-1)y
= frac11-t_1 cdot frac11-t_2
endalign$$
as expected.
answered Aug 1 at 22:19
Clement C.
46.9k33682
What is wrong is your expression for the MGF. It should be
$$
M_XY(t_1,t_2) = mathbbE[e^biglangle beginpmatrix t_1\t_2endpmatrix,beginpmatrix X\Yendpmatrix bigrangle]
= mathbbE[e^t_1X+t_2Y] tag$dagger$
$$
not $mathbbE[e^t_1X+e^t_2Y]$ (which, by linearity, would always be equal to $mathbbE[e^t_1X]+mathbbE[e^t_2Y]$—this should strike you as strange).
Using the expression from $(dagger)$, you 'll get, for $t_1,t_2<1$,
$$beginalign
M_XY(t_1,t_2) &= int_0^infty int_0^infty dxdy e^t_1x+t_2ye^-(x+y)
= int_0^infty int_0^infty dxdy e^(t_1-1)xe^(t_2-1)y\
&= int_0^infty dx e^(t_1-1)x int_0^infty dye^(t_2-1)y
= frac11-t_1 cdot frac11-t_2
endalign$$
as expected.
answered Aug 1 at 22:19
Clement C.
46.9k33682
edited Aug 1 at 22:26
answered Aug 1 at 22:19
Clement C.
46.9k33682
answered Aug 1 at 22:19
Clement C.
46.9k33682
answered Aug 1 at 22:19
Clement C.
46.9k33682
46.9k33682
add a comment |Â
add a comment |Â
up vote
0
down vote
You seek $$beginalignmathsf M_X,Y(s,t) &= mathsf E(mathrm e^sX+tY) \ &= mathsf E(mathrm e^sX~mathrm e^tY)\ &=mathsf E(mathrm e^sX)~mathsf E(mathrm e^tY) & star\ & =mathsf M_X(s)~mathsf M_Y(t)endalign$$
$star~$ Since $f_X,Y(x,y)=mathrm e^-xmathrm e^-y mathbf 1_xgeq 0mathbf 1_ygeq 0$ indicates that the random variables are independent, and infact something. Â Show this to be so, and thus use the MGF for that distribution to find the joint MGF.
answered Aug 1 at 23:36
Graham Kemp
80k43275
2
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
add a comment |Â
up vote
0
down vote
You seek $$beginalignmathsf M_X,Y(s,t) &= mathsf E(mathrm e^sX+tY) \ &= mathsf E(mathrm e^sX~mathrm e^tY)\ &=mathsf E(mathrm e^sX)~mathsf E(mathrm e^tY) & star\ & =mathsf M_X(s)~mathsf M_Y(t)endalign$$
$star~$ Since $f_X,Y(x,y)=mathrm e^-xmathrm e^-y mathbf 1_xgeq 0mathbf 1_ygeq 0$ indicates that the random variables are independent, and infact something. Â Show this to be so, and thus use the MGF for that distribution to find the joint MGF.
answered Aug 1 at 23:36
Graham Kemp
80k43275
2
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You seek $$beginalignmathsf M_X,Y(s,t) &= mathsf E(mathrm e^sX+tY) \ &= mathsf E(mathrm e^sX~mathrm e^tY)\ &=mathsf E(mathrm e^sX)~mathsf E(mathrm e^tY) & star\ & =mathsf M_X(s)~mathsf M_Y(t)endalign$$
$star~$ Since $f_X,Y(x,y)=mathrm e^-xmathrm e^-y mathbf 1_xgeq 0mathbf 1_ygeq 0$ indicates that the random variables are independent, and infact something. Â Show this to be so, and thus use the MGF for that distribution to find the joint MGF.
answered Aug 1 at 23:36
Graham Kemp
80k43275
You seek $$beginalignmathsf M_X,Y(s,t) &= mathsf E(mathrm e^sX+tY) \ &= mathsf E(mathrm e^sX~mathrm e^tY)\ &=mathsf E(mathrm e^sX)~mathsf E(mathrm e^tY) & star\ & =mathsf M_X(s)~mathsf M_Y(t)endalign$$
$star~$ Since $f_X,Y(x,y)=mathrm e^-xmathrm e^-y mathbf 1_xgeq 0mathbf 1_ygeq 0$ indicates that the random variables are independent, and infact something. Â Show this to be so, and thus use the MGF for that distribution to find the joint MGF.
answered Aug 1 at 23:36
Graham Kemp
80k43275
edited Aug 2 at 23:10
answered Aug 1 at 23:36
Graham Kemp
80k43275
answered Aug 1 at 23:36
Graham Kemp
80k43275
answered Aug 1 at 23:36
Graham Kemp
80k43275
80k43275
2
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
add a comment |Â
2
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
2
2
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
Habit from writing exponential pdf.
– Graham Kemp
Aug 2 at 23:11
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869555%2ffinding-a-joint-moment-generating-function%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
As $X$ and $Y$ are independent Exponential$(1)$ variables, the joint MGF is the product of the individual MGFs.
– StubbornAtom
Aug 2 at 15:50