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.

  • $\begingroup$ The integral doesn't diverge because of the $x_1^4 x_2^2$ unfortunately. $\endgroup$
    – Theorem
    Oct 21, 2020 at 5:24
  • $\begingroup$ There's no logic behind the divergence. Have you actually tried to evaluate the integral? If you are familiar with the conditional distributions then you can also use the law of total expectation to find this quantity. $\endgroup$ Oct 21, 2020 at 6:27

1 Answer 1

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



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

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

  1. 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)

Now let's write



  • $\begingroup$ +1 This is a fantastic answer, i did not even know the property of the second point you made. $\endgroup$
    – CCZ23
    Oct 21, 2020 at 8:12

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