Find $\mathbb{E}(X_1^4X_2^2)$ where $(X_1,X_2)$ is bivariate normal

Question

This is an exam question i just did, I want to know if my solution is correct, I guess not... If $X_i$ is bivariate normal distribution with mean $(0,0)^T$, and covariance matrix $\Sigma = \pmatrix{1 & \rho\\ \rho & 1}.$ Find the expectation of $E(X_1^4X_2^2)$.

Find the density

$$ f_X(x_1,x_2) = \frac{1}{2\pi |\Sigma|^{\frac{1}{2}}}\exp\left(x^T\Sigma^{-1}x^T \right)$$

where I think I got $\Sigma^{-1} = \frac{1}{1-\rho^2}\pmatrix{1& -\rho\\-\rho & 1}$. I then multiplied out the whole exponential, but anyway it seemed to me that the expectation would diverge since

$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x_1^4x_2^2f_X(x_1,x_2)\, dx_2 \,dx_1= \infty$$

as I cannot see the $x_1^4$ and $x_2^2$ cancelling out when I plug the infinite limits in.

Is this the right answer, or have I missed something? I can't remember the full question, but I don't think i missed anything out.

Answer 1

1. First observe that with these joint Gaussian the joint distribution can be factorized as follows

$$f_{X_1X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2|X_1}(x_2|x_1)$$

Where

$$f_{X_1}(x_1)\sim N(0;1)$$

$$f_{X_2|X_1}(x_2|x_1)\sim N(\rho x_1;1-\rho^2)$$

2. Second observe that the even moments of a standard gaussian are known and given by

$$\mathbb{E}[X^{2n}]=\frac{(2n)!}{2^n n!}$$

(the odd moments are zero)

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Now let's write

$$\mathbb{E}[X_1^4X_2^2]=\mathbb{E}[\mathbb{E}[X_1^4X_2^2|X_1]]=\mathbb{E}[X_1^4\mathbb{E}[X_2^2|X_1]]=\mathbb{E}[X_1^4(1-\rho^2+\rho^2X_1^2)]=$$

$$(1-\rho^2)\mathbb{E}[X_1^4]+\rho^2\mathbb{E}[X_1^6]=(1-\rho^2)\frac{4!}{2^2\cdot2!}+\rho^2\frac{6!}{2^3\cdot3!}=3(1-\rho^2)+15\rho^2=3(1+4\rho^2)$$