MGF and Correlation

Question

Let $X,Y$ be iid RVs with common standard normal density. Let $U=X+Y$ and $V=X^2+Y^2$. Find the MGF of the random variable $(U,V)$. Also, find the correlation coefficient between $U$ and $V$. Are $U$ and $V$ independent?

Based on sums and squares of Normal RVs, I would say that U is N(0,2) and that V is ChiSquare(2). Then the marginal densities are f(u)=$\frac 1{\sqrt {4\pi}}e^{-u^2/4}$ for $-\infty<u<\infty$ and f(v)=$\frac 12e^{-v^2/2}$ for $0<v<\infty$

Then I believe the Jacobian of this transformation is J=$\frac 1{2{\sqrt {2v-u^2}}}$ (not sure on this) which would then make the joint pdf f(u,v)=$\frac 1{\sqrt {4\pi}}e^{-v/2}(2v-u^2)^{-1/2}$

I'm not sure where to go from here or if this is even right.

Dominik · Accepted Answer · 2016-12-07T07:53:04.400

1

You don't need to calculate the joint density, you just need to calculate the moment generating function. Since $X$ and $Y$ are independent, the MGF factorizes in the following way:

$$E[e^{t_1U + t_2V}] = E[e^{(t_1 X + t_2 X^2) + (t_1 Y + t_2 Y^2)}] = E[e^{t_1 X + t_2 X^2}] E[e^{t_1 Y + t_2 Y^2}]$$

Now you just need to calculate the integral $$\begin{align*} &\frac{1}{\sqrt{2\pi}} \int \limits_{-\infty}^\infty \exp(t_1x + t_2x^2 - x^2/2) \, dx \\ ={}& \frac{1}{\sqrt{2\pi}} \int \limits_{-\infty}^\infty \exp\left((t_2 - 1/2)\left(x + \frac{t_1}{2 \sqrt{t_2 - 1/2}}\right)^2 - \frac{t_1^2}{4t_2 - 2} \right) \, dx \\ ={}& \frac{1}{\sqrt{1 - 2t_2}} \exp\left(\frac{t_1^2}{2 - 4t_2}\right) \end{align*}$$

(The integral diverges for $t_2 \ge \frac{1}{2}$).

To get the mixed moments of $U$ and $V$, you need to consider the connection between the derivative of the MGF and the mixed moments. As for the independence, you just need to check if the moment generating function factorizes in the way that it does with independent random variables.

edited Dec 07 '16 at 07:53

answered Dec 03 '16 at 15:56

Dominik

20,241

That integral you wrote is the MGF of (U,V)? Then for independence I would first see if the correlation coefficient is nonzero. – user395073 Dec 03 '16 at 16:55
1

This is not yet the MGF. Try to understand the steps that I have written and where the integrals come from. – Dominik Dec 03 '16 at 16:56
So just to make sure I'm following: you factorized the MGF, then reordered the exponents to group X and Y terms. The integral then would be the first expectation, since that is the e term multiplied by the pdf of a standard normal X, integrated across the real line. – user395073 Dec 03 '16 at 18:29
That is correct. – Dominik Dec 04 '16 at 00:53
Ok so by doing the same for Y, I get that the joint MGF is $\frac 1{1-2t_2}*e^{({\frac {t_1^2}{1-2t_2}})}$ – user395073 Dec 04 '16 at 02:56
Could you confirm this? (Sorry if my MathJax is not so good yet) – user395073 Dec 04 '16 at 02:57
I get the same result – Dominik Dec 04 '16 at 10:22
Perfect. Then I obtained that U and V are uncorrelated but not independent. – user395073 Dec 04 '16 at 14:39
Quick question: how did you get the square root in the integral above? – user395073 Dec 05 '16 at 01:50
That's from the linear substitution. – Dominik Dec 05 '16 at 08:26
In the second integral there, where did the - sign come from between x and the t1 term? – user395073 Dec 07 '16 at 04:20
My bad, that should be a plus. Thanks, I've fixed it now. – Dominik Dec 07 '16 at 07:53

score 0 · Answer 2 · answered Feb 09 '25 at 09:17

There is a general formula for gaussian integral $$\int_{-\infty}^{\infty}e^{-a x^2 + b x }\,dx=\sqrt{\frac{\pi}{a}}\,e^{\frac{b^2}{4a}}$$

Now, let's calculate the joint MGF

$$ M_{U,V}(t)=E(e^{t_1U+t_2V})=E(e^{t_1X+t_2X^2})E(e^{t_1Y+t_2Y^2}) $$

Since X and Y have identical distributions $E(e^{t_1X+t_2X^2})=E(e^{t_1Y+t_2Y^2})$, and we have

$$ M_{U,V}(t) = E^2(e^{t_1X+t_2X^2}) = (\int_{-\infty}^{\infty}e^{t_1x+t_2x^2}\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}dx)^2 = (\int_{-\infty}^{\infty}\frac{e^{-(1/2-t_2)x^2+t_1x}}{\sqrt{2\pi}}dx)^2 $$

Let $a=1/2-t_2$ and $b=t_1$

$$ M_{U,V}(t) = (\frac{1}{\sqrt{2\pi}}\sqrt{\frac{\pi}{1/2-t_2}}\,e^{\frac{t_1^2}{4(1/2-t_2)}})^2 = \frac{e^{\frac{t_1^2}{1-2t_2}}}{1-2t_2}$$

Correlation Coefficient

We know that $U\sim N(0,2)$ and $V\sim \chi^2_2$. The only thing we need for correlation coefficient is $E(UV)$

$$ E(UV) = E(X^3+XY^2+YX^2+Y^3) = E(X^3+Y^3) $$

Since X and Y have standard normal distributions we have $E(X^3)=E(Y^3)=0$ (See this) which means that $E(UV)=0$ and

$$Cov(U,V)=E(U,V)-E(U)E(V)=0-0×2=0$$

which means

$$ Corr(U,V)=\frac{Cov(U,V)}{\sqrt{Var(U)Var(V)}} = 0$$

MGF and Correlation

2 Answers2

Correlation Coefficient