Finding $E[X_{(1)}]$ when there are exactly two observations.

Question

I am trying to calculate $E[X_{(1)}]$ given that $X_1,X_2\sim_{iid} N(0,\sigma^2)$.

My first response is to use the order statistics distribution

$$f_{X_{(1)}}(x)=2f(x)[1-F(x)]$$

but if I try to use this straight forwardly I would have to evaluate

$$\int_{-\infty}^{\infty} \frac{2x}{\sqrt{2\pi}\sigma}e^{-x^2/\sigma^2} \left[ \int_{-\infty}^{x}\frac{1}{\sqrt{2\pi }\sigma}e^{-t^2/\sigma^2} dt \right]dx$$

and I do not have the proper skill to evaluate this.

My notes tell me, on the other hand, that you can evaluate this expectation using a joint pdf where

$$f_{X_1,X_2}(x_1,x_2)=nf_{X_1}(x_1)f_{X_2}(x_2) \quad \text{where} \quad -\infty < x_1<x_2<\infty$$.

I believe that I have a lack of understanding of joint distributions . . .

I see that because $X_{(1)}<X_{(2)}$ the support of the joint is $-\infty < x_1 < x_2 < \infty$ but I do not know where the coefficient "$n$" comes from.

I appreciate your help.

Answer 1

For simplicity, take $\sigma=1$; the answer for general $\sigma$ will be $\sigma$ times as large. Break the integral $\int 2xf(x)[1-F(x)]\,dx$ into two pieces, the first of which is zero. The second piece is handled using integration by parts: Write $$-2\int xf(x)F(x)\,dx=2\int u\,dv\tag1$$ where $u=F(x)$ and $v=f(x)$, so that $du=f(x)$ and $dv=-xf(x)\,dx$, and therefore $$ 2\int u\,dv = 2uv -2\int v\,du.\tag2 $$ Evaluating $uv$ from $-\infty$ to $\infty$ will give zero, whereas the second term on the RHS of (2) evaluates to $$-2\int_{-\infty}^\infty v\,du=-2\int_{-\infty}^\infty f(x)^2\,dx=-\frac2{2\pi}\int_{-\infty}^\infty e^{-x^2}dx.\tag3 $$ That last integral equals $\sqrt \pi$, so your final answer (for the $\sigma=1$ case) is $-\frac 1{\sqrt\pi}$, and for the general case it's $-\frac\sigma{\sqrt\pi}$.

Answer 2

The first formula for $f_{X_(1)}$ is correct. There is no $n$ in the last formula. It must be a typo.
Direct evaluation gives $2\int_{-\infty}^{t} f(x_1) [1-F(x_1)] dx_1$. You cannot evaluate this explicitly but you can differentiate and get the formula for the density.

To evaluate the expectation you have to find $\int 2xf(x)[1-F(x)]dx$. Note that $xe^{-{x^{2} }/2}$ is the derivative of $-e^{-{x^{2}} /2}$. Integration by parts gives you the expectation of $X_{(1)}$ easily.