4
$\begingroup$

What is the distribution of the following monomial? $$X^a \cdot Y^b$$ where $X$ and $Y$ are normal random variables and $a$ and $b$ are natural numbers.

For example, when $X \sim N(0,1)$, $a=2$, and $b=0$ it is a Chi-squared distribution, which has a variance of 2.

What if we have $n$ independent variables $X_1, X_2, \dots , X_n$, with $X_i \sim N(0,\sigma^2)$ and some natural numbers $p_1, p_2, \dots,p_n$. What can we say about the variance of the following r.v.?

$$X_1^{p_1} \cdot X_2^{p_2} \cdots X_n^{p_n}$$

$\endgroup$
  • 1
    $\begingroup$ Of possible interest: link. $\endgroup$ – user10525 May 11 '12 at 8:28
5
$\begingroup$

The first question about a distribution has no convenient general answer, because AFAIK nobody has assigned names to such distributions nor extensively studied and characterized them except when both $a$ and $b$ are $2$ or less.

Concerning the second question about the variances, as shorthand write $\mathbf{p}=(p_1,p_2,\ldots,p_n)$ and $\mathbf{x^p} = X_1^{p_1} X_2^{p_2} \cdots X_n^{p_n}$.

Recall that for the standard Normal distribution the $k^\text{th}$ moment is $0$ when $k$ is odd and otherwise equals $(k-1)!! = (k-1)(k-3)\cdots(3)(1)$. From this and the independence assumption, the expression for the expectation of $\mathbf{x^p}$ is immediate: it equals $0$ when one or more of the $p_i$ is odd and otherwise is the product of the $(p_i-1)!!$, which I will similarly abbreviate $(\mathbf{p-1})!!$.

By definition,

$$\text{Var}(\mathbf{x^p}) = \mathbb{E}[(\mathbf{x^p})^2] - \mathbb{E}[\mathbf{x^p}]^2 = (2\mathbf{p}\mathbf{-1})!! - ((\mathbf{p-1})!!)^2.$$

When the variables are scaled to have variance $\sigma^2$, $\mathbf{x^p}$ will be multiplied by $|\sigma|^{p_1+p_2+\cdots+p_n}$, whence its variance will be multiplied by $\sigma^{2(p_1+p_2+\cdots+p_n)}$.

Example

Let $\mathbf{p} = (2,4)$:

  • $(2\mathbf{p}\mathbf{-1})!! = 3!! 7!! = [3(1)][7(5)(3)(1)] = 315$;

  • $((\mathbf{p-1})!!)^2 = 1!!3!! = ([1][3(1)])^2 = 9$;

  • $\text{Var}(X_1^2 X_2^4) = (315 - 9)\sigma^{2(2+4)} = 306\sigma^{12}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.