We sample $m$ standard Gaussian random variables $x_i \sim N(0,1)$. Then we select the $n \leq m$ smallest ones in absolute value. Compute the expectation of $X:=\sum_{i=1}^n x_i^2$.
I know that since $x_i$ is a standard Gaussian random variable $X$ is chi-squared distributed with $n$ degrees of freedom. The expectation of a chi-squared random variable is the number of degrees of freedom, so in this case $n$. Since we are first sampling $m$ variables $x_i \sim N(0,1)$ and then choosing the $n$ smallest ones we obtain that $E[X]\leq n$, but what is it exactly?
Thank you very much for your help.
Ok: So I want to compute the first, second, …, n-th order of statistic of a chi-squared random variable with 1 degree of freedom, compute the expected values and add them together.
The cumulative distribution function of a chi-squared random variable with 1 degree of freedom is given by $F(x) = Erf(\sqrt{\frac{x}{2}})$ where $Erf$ denotes the Gaussian error function and the density function is $f(x)=\frac{1}{\sqrt{2\pi x}}e^{-\frac{x}{2}}$ Hence, the density function is
$f_{X_{(r)}}(x) = \frac{m!}{(r-1)!(m-r)!} f(x)[F(x)]^{r-1}[1-F(x)]^{m-r}$.
The expectation of the r-th order of statistics is given by $\int_0^\infty xf(x) dx = \frac{m!}{(r-1)!(m-r)!} \frac{1}{\sqrt{2\pi}} \int_0^\infty \sqrt{x}e^{-x/2}(erf({\sqrt{\frac{x}{2}}}))^{r-1} (1-erf(\sqrt{\frac{x}{2}}))^{m-r}dx$
I am having problems computing this term, does someone have a further tip for me?
EDIT: The integral seems to be impossible to compute for a general $m$, although it is possible if one plugs in a specific value. Hence, I am now closing the question.
Best Answer
It would be the expected value of the 1st order statistic, 2nd order statistic, up to nth order statistic of a chi squared random variable with 1 degree of freedom from a sample of size m, added together.