Let $1<p<2.$ Let $G$ be a complex Gaussian random variable. then what is the value of $\mathbb{E}[|G|^p]$ ? The symbol $\mathbb{E}$ denotes the expectation of a random variable.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Let $X:=|G|^2=U^2+V^2$, where $U:=\Re G$ and $V:=\Im G$, so that $U$ and $V$ are iid standard normal random variables (r.v.'s). Then $X$ has the chi-squared distribution with 2 degrees of freedom, which is the same as the gamma distribution with parameters $1,2$. So, for all real $p>-2$, $$E|G|^p=EX^{p/2}=\int_0^\infty x^{p/2}\frac12\,e^{-x/2}dx=2^{p/2}\Gamma(p/2+1).$$
answered Mar 16, 2018 at 1:45
-
$\begingroup$ What if $G$ is standard complex Gaussian? i.e $U,V$ are i.i.d. and follows normal distribution with mean zero and variance $\frac{1}{2}..$ $\endgroup$– MathbuffCommented Mar 19, 2018 at 16:51
-
$\begingroup$ @Mathbuff : Then "your" $X$ will be $1/2$ of "my" $X$, and so, the result $2^{p/2}\Gamma(p/2+1)$ will get multiplied by $(1/2)^{p/2}$, giving simply $\Gamma(p/2+1)$ in the answer. $\endgroup$ Commented Mar 19, 2018 at 22:16
-
$\begingroup$ @ Iosif Pinelis . But for $p=2$ it is not matching. Since $\mathbb{E}|G|^2=1$ for $G$ standard complex Gaussian random variable!! $\endgroup$– MathbuffCommented Mar 20, 2018 at 9:50
-
$\begingroup$ @Mathbuff : It is matching: for $p=2$ we have $E|G|^2=\Gamma(p/2+1)=\Gamma(2)=1$. $\endgroup$ Commented Mar 20, 2018 at 13:53
-
$\begingroup$ @ Iosif Pinelis . Yea . sorry. I took $\Gamma(n)=n!.$! Now I got it. $\endgroup$– MathbuffCommented Mar 20, 2018 at 14:47