Computing the joint moment generating function for two functions of two random variables
Let $X$ and $Y$ be i.i.d random variables in the plane with a pdf
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2}/2) hspace{1cm}
-infty < x< infty.$$
Let $U = X + Y$ and $V = X^{2} + Y^{2}$. Compute the joint mgf of $U$
and $V$.
My first step was to compute the MGF of $U$ as follows:
begin{align*}
mathbb{E}[e^{sU}] = mathbb{E}[e^{s(X + Y)}] = mathbb{E}[e^{sX}e^{sY}] \
= mathbb{E}[e^{sX}] cdot mathbb{E}[e^{sY}] \
= M_{X}(s) cdot M_{Y}(s) \
= (e^{s^2/2})^{2} \
= e^{s^{2}},
end{align*}
where $t in mathbb{R}$. Note that we can separate expectations due to independence.
Now I can compute the MGF of $V$ as follows:
begin{align*}
mathbb{E}[e^{tV}] = mathbb{E}[e^{t(X^{2} + Y^{2})}] = mathbb{E}[e^{tX^{2}}] cdot mathbb{E}[e^{tY^{2}}] \
= M_{X^{2}}(t) cdot M_{Y^{2}}(t),
end{align*}
To proceed, I need the mgf of $X^{2}$ and $Y^{2}$. So, maybe is there an easier way to approach this problem? If not, how would I proceed. Then, at the end, I can just multiply the two mgf I found individually? I'm not sure if this approach is correct though, as it would assume $U$ and $V$ to be independent.
The joint moment generating function of $U$ and $V$ is given by
$$mathbb{E}[e^{sU + tV}] $$
probability probability-theory probability-distributions moment-generating-functions
asked Dec 12 '18 at 21:43
josephjoseph
46210
add a comment |
Let $X$ and $Y$ be i.i.d random variables in the plane with a pdf
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2}/2) hspace{1cm}
-infty < x< infty.$$
Let $U = X + Y$ and $V = X^{2} + Y^{2}$. Compute the joint mgf of $U$
and $V$.
My first step was to compute the MGF of $U$ as follows:
begin{align*}
mathbb{E}[e^{sU}] = mathbb{E}[e^{s(X + Y)}] = mathbb{E}[e^{sX}e^{sY}] \
= mathbb{E}[e^{sX}] cdot mathbb{E}[e^{sY}] \
= M_{X}(s) cdot M_{Y}(s) \
= (e^{s^2/2})^{2} \
= e^{s^{2}},
end{align*}
where $t in mathbb{R}$. Note that we can separate expectations due to independence.
Now I can compute the MGF of $V$ as follows:
begin{align*}
mathbb{E}[e^{tV}] = mathbb{E}[e^{t(X^{2} + Y^{2})}] = mathbb{E}[e^{tX^{2}}] cdot mathbb{E}[e^{tY^{2}}] \
= M_{X^{2}}(t) cdot M_{Y^{2}}(t),
end{align*}
To proceed, I need the mgf of $X^{2}$ and $Y^{2}$. So, maybe is there an easier way to approach this problem? If not, how would I proceed. Then, at the end, I can just multiply the two mgf I found individually? I'm not sure if this approach is correct though, as it would assume $U$ and $V$ to be independent.
The joint moment generating function of $U$ and $V$ is given by
$$mathbb{E}[e^{sU + tV}] $$
probability probability-theory probability-distributions moment-generating-functions
asked Dec 12 '18 at 21:43
josephjoseph
46210
Why don't you write down the definition of the joint MGF of $U$ and $V$ and start from there? As you noted at the end, computing the individual MGFs might not help with the joint MGF since $U$ and $V$ are not independent.
– angryavian
Dec 12 '18 at 21:45
I tried starting with that approach. I edited my post with the joint moment generating function. I don't know how to proceed from just that, though. Maybe conditioning?
– joseph
Dec 12 '18 at 21:47
It looks like $X$ and $Y$ are standard normal, and then $X^2,Y^2$ are chi-squared with one degree of freedom.
– gd1035
Dec 12 '18 at 21:58
Yeah I saw that on wikipedia but I have not learned chi-square distribution.
– joseph
Dec 12 '18 at 21:59
add a comment |
Let $X$ and $Y$ be i.i.d random variables in the plane with a pdf
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2}/2) hspace{1cm}
-infty < x< infty.$$
Let $U = X + Y$ and $V = X^{2} + Y^{2}$. Compute the joint mgf of $U$
and $V$.
My first step was to compute the MGF of $U$ as follows:
begin{align*}
mathbb{E}[e^{sU}] = mathbb{E}[e^{s(X + Y)}] = mathbb{E}[e^{sX}e^{sY}] \
= mathbb{E}[e^{sX}] cdot mathbb{E}[e^{sY}] \
= M_{X}(s) cdot M_{Y}(s) \
= (e^{s^2/2})^{2} \
= e^{s^{2}},
end{align*}
where $t in mathbb{R}$. Note that we can separate expectations due to independence.
Now I can compute the MGF of $V$ as follows:
begin{align*}
mathbb{E}[e^{tV}] = mathbb{E}[e^{t(X^{2} + Y^{2})}] = mathbb{E}[e^{tX^{2}}] cdot mathbb{E}[e^{tY^{2}}] \
= M_{X^{2}}(t) cdot M_{Y^{2}}(t),
end{align*}
To proceed, I need the mgf of $X^{2}$ and $Y^{2}$. So, maybe is there an easier way to approach this problem? If not, how would I proceed. Then, at the end, I can just multiply the two mgf I found individually? I'm not sure if this approach is correct though, as it would assume $U$ and $V$ to be independent.
The joint moment generating function of $U$ and $V$ is given by
$$mathbb{E}[e^{sU + tV}] $$
probability probability-theory probability-distributions moment-generating-functions
asked Dec 12 '18 at 21:43
josephjoseph
46210
Let $X$ and $Y$ be i.i.d random variables in the plane with a pdf
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2}/2) hspace{1cm}
-infty < x< infty.$$
Let $U = X + Y$ and $V = X^{2} + Y^{2}$. Compute the joint mgf of $U$
and $V$.
My first step was to compute the MGF of $U$ as follows:
begin{align*}
mathbb{E}[e^{sU}] = mathbb{E}[e^{s(X + Y)}] = mathbb{E}[e^{sX}e^{sY}] \
= mathbb{E}[e^{sX}] cdot mathbb{E}[e^{sY}] \
= M_{X}(s) cdot M_{Y}(s) \
= (e^{s^2/2})^{2} \
= e^{s^{2}},
end{align*}
where $t in mathbb{R}$. Note that we can separate expectations due to independence.
Now I can compute the MGF of $V$ as follows:
begin{align*}
mathbb{E}[e^{tV}] = mathbb{E}[e^{t(X^{2} + Y^{2})}] = mathbb{E}[e^{tX^{2}}] cdot mathbb{E}[e^{tY^{2}}] \
= M_{X^{2}}(t) cdot M_{Y^{2}}(t),
end{align*}
To proceed, I need the mgf of $X^{2}$ and $Y^{2}$. So, maybe is there an easier way to approach this problem? If not, how would I proceed. Then, at the end, I can just multiply the two mgf I found individually? I'm not sure if this approach is correct though, as it would assume $U$ and $V$ to be independent.
The joint moment generating function of $U$ and $V$ is given by
$$mathbb{E}[e^{sU + tV}] $$
probability probability-theory probability-distributions moment-generating-functions
probability probability-theory probability-distributions moment-generating-functions
asked Dec 12 '18 at 21:43
josephjoseph
46210
asked Dec 12 '18 at 21:43
josephjoseph
46210
edited Dec 12 '18 at 21:47
joseph
asked Dec 12 '18 at 21:43
josephjoseph
46210
asked Dec 12 '18 at 21:43
josephjoseph
46210
asked Dec 12 '18 at 21:43
josephjoseph
46210
46210
Why don't you write down the definition of the joint MGF of $U$ and $V$ and start from there? As you noted at the end, computing the individual MGFs might not help with the joint MGF since $U$ and $V$ are not independent.
– angryavian
Dec 12 '18 at 21:45
I tried starting with that approach. I edited my post with the joint moment generating function. I don't know how to proceed from just that, though. Maybe conditioning?
– joseph
Dec 12 '18 at 21:47
It looks like $X$ and $Y$ are standard normal, and then $X^2,Y^2$ are chi-squared with one degree of freedom.
– gd1035
Dec 12 '18 at 21:58
Yeah I saw that on wikipedia but I have not learned chi-square distribution.
– joseph
Dec 12 '18 at 21:59
add a comment |
Why don't you write down the definition of the joint MGF of $U$ and $V$ and start from there? As you noted at the end, computing the individual MGFs might not help with the joint MGF since $U$ and $V$ are not independent.
– angryavian
Dec 12 '18 at 21:45
I tried starting with that approach. I edited my post with the joint moment generating function. I don't know how to proceed from just that, though. Maybe conditioning?
– joseph
Dec 12 '18 at 21:47
It looks like $X$ and $Y$ are standard normal, and then $X^2,Y^2$ are chi-squared with one degree of freedom.
– gd1035
Dec 12 '18 at 21:58
Yeah I saw that on wikipedia but I have not learned chi-square distribution.
– joseph
Dec 12 '18 at 21:59
Why don't you write down the definition of the joint MGF of $U$ and $V$ and start from there? As you noted at the end, computing the individual MGFs might not help with the joint MGF since $U$ and $V$ are not independent.
– angryavian
Dec 12 '18 at 21:45
Why don't you write down the definition of the joint MGF of $U$ and $V$ and start from there? As you noted at the end, computing the individual MGFs might not help with the joint MGF since $U$ and $V$ are not independent.
– angryavian
Dec 12 '18 at 21:45
I tried starting with that approach. I edited my post with the joint moment generating function. I don't know how to proceed from just that, though. Maybe conditioning?
– joseph
Dec 12 '18 at 21:47
I tried starting with that approach. I edited my post with the joint moment generating function. I don't know how to proceed from just that, though. Maybe conditioning?
– joseph
Dec 12 '18 at 21:47
It looks like $X$ and $Y$ are standard normal, and then $X^2,Y^2$ are chi-squared with one degree of freedom.
– gd1035
Dec 12 '18 at 21:58
It looks like $X$ and $Y$ are standard normal, and then $X^2,Y^2$ are chi-squared with one degree of freedom.
– gd1035
Dec 12 '18 at 21:58
Yeah I saw that on wikipedia but I have not learned chi-square distribution.
– joseph
Dec 12 '18 at 21:59
Yeah I saw that on wikipedia but I have not learned chi-square distribution.
– joseph
Dec 12 '18 at 21:59
add a comment |
1 Answer
1
active
oldest
votes
You have written down the joint MGF correctly. Now just plug in the definitions of $U$ and $V$, and keep going, using independence where you can to simplify things.
$$E[e^{sU + tV}] = E[e^{sX+tX^2 + sY + tY^2}] = E[e^{sX + tX^2}] E[e^{sY + tY^2}] = E[e^{sX + tX^2}]^2$$
To compute the last term, I believe you can compute the integral directly, i.e.
$$E[e^{sX+tX^2}] = int_{-infty}^infty e^{sx+tx^2} f(x) , dx = frac{1}{sqrt{2pi}} int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx.$$
You might need to do some casework (e.g., when $t=1/2$ this integral diverges, etc.).
Edit:
When $t ge 1/2$, I believe the integral $int_{-infty}^infty e^{sx + (t - frac{1}{2}) x^2} , dx$ diverges.
When $t < 1/2$ we can write
$$sx + tx^2 - x^2/2 = -frac{(x-mu)^2}{2sigma^2} + c$$
where $mu = frac{s}{1-2t}$ and $sigma^2 = frac{1}{1-2t}$ and $c = frac{s^2}{2(1-2t)}$.
So the integral is
$$frac{1}{sqrt{2pi}}int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx = frac{1}{sqrt{2pi}} e^c int_{-infty}^infty e^{-frac{(x-mu)^2}{2sigma^2}} , dx
= frac{1}{sqrt{2pi}} e^c sqrt{2 pi sigma^2} = frac{1}{sqrt{1-2t}} expleft(frac{s^2}{2(1-2t)}right).$$
It seems like the final answer is thus
$$frac{1}{1-2t} expleft(frac{s^2}{1-2t}right), qquad t < 1/2$$
but I may have made some typos.
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
add a comment |
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You have written down the joint MGF correctly. Now just plug in the definitions of $U$ and $V$, and keep going, using independence where you can to simplify things.
$$E[e^{sU + tV}] = E[e^{sX+tX^2 + sY + tY^2}] = E[e^{sX + tX^2}] E[e^{sY + tY^2}] = E[e^{sX + tX^2}]^2$$
To compute the last term, I believe you can compute the integral directly, i.e.
$$E[e^{sX+tX^2}] = int_{-infty}^infty e^{sx+tx^2} f(x) , dx = frac{1}{sqrt{2pi}} int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx.$$
You might need to do some casework (e.g., when $t=1/2$ this integral diverges, etc.).
Edit:
When $t ge 1/2$, I believe the integral $int_{-infty}^infty e^{sx + (t - frac{1}{2}) x^2} , dx$ diverges.
When $t < 1/2$ we can write
$$sx + tx^2 - x^2/2 = -frac{(x-mu)^2}{2sigma^2} + c$$
where $mu = frac{s}{1-2t}$ and $sigma^2 = frac{1}{1-2t}$ and $c = frac{s^2}{2(1-2t)}$.
So the integral is
$$frac{1}{sqrt{2pi}}int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx = frac{1}{sqrt{2pi}} e^c int_{-infty}^infty e^{-frac{(x-mu)^2}{2sigma^2}} , dx
= frac{1}{sqrt{2pi}} e^c sqrt{2 pi sigma^2} = frac{1}{sqrt{1-2t}} expleft(frac{s^2}{2(1-2t)}right).$$
It seems like the final answer is thus
$$frac{1}{1-2t} expleft(frac{s^2}{1-2t}right), qquad t < 1/2$$
but I may have made some typos.
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
add a comment |
You have written down the joint MGF correctly. Now just plug in the definitions of $U$ and $V$, and keep going, using independence where you can to simplify things.
$$E[e^{sU + tV}] = E[e^{sX+tX^2 + sY + tY^2}] = E[e^{sX + tX^2}] E[e^{sY + tY^2}] = E[e^{sX + tX^2}]^2$$
To compute the last term, I believe you can compute the integral directly, i.e.
$$E[e^{sX+tX^2}] = int_{-infty}^infty e^{sx+tx^2} f(x) , dx = frac{1}{sqrt{2pi}} int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx.$$
You might need to do some casework (e.g., when $t=1/2$ this integral diverges, etc.).
Edit:
When $t ge 1/2$, I believe the integral $int_{-infty}^infty e^{sx + (t - frac{1}{2}) x^2} , dx$ diverges.
When $t < 1/2$ we can write
$$sx + tx^2 - x^2/2 = -frac{(x-mu)^2}{2sigma^2} + c$$
where $mu = frac{s}{1-2t}$ and $sigma^2 = frac{1}{1-2t}$ and $c = frac{s^2}{2(1-2t)}$.
So the integral is
$$frac{1}{sqrt{2pi}}int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx = frac{1}{sqrt{2pi}} e^c int_{-infty}^infty e^{-frac{(x-mu)^2}{2sigma^2}} , dx
= frac{1}{sqrt{2pi}} e^c sqrt{2 pi sigma^2} = frac{1}{sqrt{1-2t}} expleft(frac{s^2}{2(1-2t)}right).$$
It seems like the final answer is thus
$$frac{1}{1-2t} expleft(frac{s^2}{1-2t}right), qquad t < 1/2$$
but I may have made some typos.
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
add a comment |
You have written down the joint MGF correctly. Now just plug in the definitions of $U$ and $V$, and keep going, using independence where you can to simplify things.
$$E[e^{sU + tV}] = E[e^{sX+tX^2 + sY + tY^2}] = E[e^{sX + tX^2}] E[e^{sY + tY^2}] = E[e^{sX + tX^2}]^2$$
To compute the last term, I believe you can compute the integral directly, i.e.
$$E[e^{sX+tX^2}] = int_{-infty}^infty e^{sx+tx^2} f(x) , dx = frac{1}{sqrt{2pi}} int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx.$$
You might need to do some casework (e.g., when $t=1/2$ this integral diverges, etc.).
Edit:
When $t ge 1/2$, I believe the integral $int_{-infty}^infty e^{sx + (t - frac{1}{2}) x^2} , dx$ diverges.
When $t < 1/2$ we can write
$$sx + tx^2 - x^2/2 = -frac{(x-mu)^2}{2sigma^2} + c$$
where $mu = frac{s}{1-2t}$ and $sigma^2 = frac{1}{1-2t}$ and $c = frac{s^2}{2(1-2t)}$.
So the integral is
$$frac{1}{sqrt{2pi}}int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx = frac{1}{sqrt{2pi}} e^c int_{-infty}^infty e^{-frac{(x-mu)^2}{2sigma^2}} , dx
= frac{1}{sqrt{2pi}} e^c sqrt{2 pi sigma^2} = frac{1}{sqrt{1-2t}} expleft(frac{s^2}{2(1-2t)}right).$$
It seems like the final answer is thus
$$frac{1}{1-2t} expleft(frac{s^2}{1-2t}right), qquad t < 1/2$$
but I may have made some typos.
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
You have written down the joint MGF correctly. Now just plug in the definitions of $U$ and $V$, and keep going, using independence where you can to simplify things.
$$E[e^{sU + tV}] = E[e^{sX+tX^2 + sY + tY^2}] = E[e^{sX + tX^2}] E[e^{sY + tY^2}] = E[e^{sX + tX^2}]^2$$
To compute the last term, I believe you can compute the integral directly, i.e.
$$E[e^{sX+tX^2}] = int_{-infty}^infty e^{sx+tx^2} f(x) , dx = frac{1}{sqrt{2pi}} int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx.$$
You might need to do some casework (e.g., when $t=1/2$ this integral diverges, etc.).
Edit:
When $t ge 1/2$, I believe the integral $int_{-infty}^infty e^{sx + (t - frac{1}{2}) x^2} , dx$ diverges.
When $t < 1/2$ we can write
$$sx + tx^2 - x^2/2 = -frac{(x-mu)^2}{2sigma^2} + c$$
where $mu = frac{s}{1-2t}$ and $sigma^2 = frac{1}{1-2t}$ and $c = frac{s^2}{2(1-2t)}$.
So the integral is
$$frac{1}{sqrt{2pi}}int_{-infty}^infty e^{sx + tx^2 - x^2/2} , dx = frac{1}{sqrt{2pi}} e^c int_{-infty}^infty e^{-frac{(x-mu)^2}{2sigma^2}} , dx
= frac{1}{sqrt{2pi}} e^c sqrt{2 pi sigma^2} = frac{1}{sqrt{1-2t}} expleft(frac{s^2}{2(1-2t)}right).$$
It seems like the final answer is thus
$$frac{1}{1-2t} expleft(frac{s^2}{1-2t}right), qquad t < 1/2$$
but I may have made some typos.
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
edited Dec 13 '18 at 0:26
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
answered Dec 12 '18 at 21:49
angryavianangryavian
39.3k23280
39.3k23280
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
add a comment |
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
How come you were able to interchange the $s$ and the $t$'s?
– joseph
Dec 12 '18 at 21:51
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
@joseph Sorry about that mistake.
– angryavian
Dec 12 '18 at 22:03
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
how can I solve that integral? it looks challenging
– joseph
Dec 12 '18 at 22:10
add a comment |
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Why don't you write down the definition of the joint MGF of $U$ and $V$ and start from there? As you noted at the end, computing the individual MGFs might not help with the joint MGF since $U$ and $V$ are not independent.
– angryavian
Dec 12 '18 at 21:45
I tried starting with that approach. I edited my post with the joint moment generating function. I don't know how to proceed from just that, though. Maybe conditioning?
– joseph
Dec 12 '18 at 21:47
It looks like $X$ and $Y$ are standard normal, and then $X^2,Y^2$ are chi-squared with one degree of freedom.
– gd1035
Dec 12 '18 at 21:58
Yeah I saw that on wikipedia but I have not learned chi-square distribution.
– joseph
Dec 12 '18 at 21:59