estimates Gaussian moments

Question

Let $X_i \sim N(0,\sigma_i^2)$. Let $k\geq0$ be a fixed integer. I would like to compute $$A:=E[|X_1-X_2|^k|X_2|^k]$$

My idea was \begin{align*} A=&\int_{\mathbb{R}^2}|x_1-x_2|^k |x_2|^k e^{-\frac{1}{2\sigma_1^2}|x_1|^2}e^{-\frac{1}{2\sigma_2^2}|x_2|^2}dx_1dx_2\\ =&\int_{\mathbb{R}^2}|z_1|^k |z_2|^k e^{-\frac{1}{2\sigma_1^2}|z_1+z_2|^2}e^{-\frac{1}{2\sigma_2^2}|z_2|^2}dz_1dz_2\\ &= \int_{\mathbb{R}}|z_2|^k e^{-\frac{1}{2\sigma_2^2}|z_2|^2} E[|Z_1|^k]dz_2 \end{align*} where $Z_1\sim N(-z_2,\sigma_1^2)$. Using Kummer's confluent hypergeometric fuction (denoted by $M(a,b,z)$) we have that the non-central moments of a r.v. $X\sim N(\mu,\sigma^2)$ are: $$E[|X|^k] = \sigma^k 2^{k/2} \frac{\Gamma \left( \frac{ k+1 }{2} \right)}{\sqrt{\pi}}M(-\frac{k}{2},\frac{1}{2},-\frac{\mu^2}{2\sigma^2}).$$ Hence \begin{align*} A=& \int_{\mathbb{R}}|z_2|^k e^{-\frac{1}{2\sigma_2^2}|z_2|^2} E[|Z_1|^k]dz_2\\ =& C_k \sigma_1^k \int_{\mathbb{R}}|z_2|^k e^{-\frac{1}{2\sigma_2^2}|z_2|^2}M(-\frac{k}{2},\frac{1}{2},-\frac{z_2^2}{2\sigma_1^2})dz_2 \end{align*} where $C_k$ is the constant depending on $k$. If $M$ happens to be bounded for this case. Then I can get the following bound \begin{align*} A=& C_k \sigma_1^k \int_{\mathbb{R}}|z_2|^k e^{-\frac{1}{2\sigma_2^2}|z_2|^2}M(-\frac{k}{2},\frac{1}{2},-\frac{z_2^2}{2\sigma_1^2})dz_2\\ &\leq \widetilde{C}_k\sigma_1^k \sigma_2^k. \end{align*}

Is this procedure correct? If not, does anyone have any idea on how to find bound for $A$ depending on $\sigma_1,\sigma_2$?

Thanks a lot!

Answer 1

Question: Let $X_i \sim N(0,\sigma_i^2)$. I would like to compute $E[|X_1-X_2|^k|X_2|^k]$

Bounds shmounds¡? For given values of integer $k$, it is possible to obtain exact closed-form solutions. In particular, if $X_1$ and $X_2$ are independent (which appears to be your intention), the joint pdf of $(X_1,X_2)$, say $f(x_1,x_2)$, is:

_{(source: tri.org.au)}

Then, $E\big[\,\big|X_1-X_2\big|^k \,\big|X_2\big|^k \, \big]$ can be derived with:

_{(source: tri.org.au)}

where I am using the Expect function from the mathStatica package for Mathematica to do the nitty-gritties.

For example, the solution when $k = 2$ is:

_{(source: tri.org.au)}

Here are the solutions corresponding to $k = 1$ to 6:

_{(source: tri.org.au)}

There is no problem deriving higher 'moments', if required.

Notes

1. If a general solution is required for arbitrary $k$, it may perhaps be possible to obtain same via induction, treating the even and odd $k$ cases separately.

2. As disclosure, I should perhaps add that I am one of the authors of the software function used above.