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Let $X_{(1)}\leq X_{(2)}$ be the order statistics for a random sample of size $2$ from a normal distribution with mean $\mu$ and variance $\sigma ^{2}$.

Evaluate $E(X_{(1)})$, $E(X_{(2)})$, $\mathrm{Var}(X{(1)})$, $\mathrm{Var}(X{(2)})$ and $\mathrm{Cov}(X{(1)},X_{(2)})$.

My attempt: In general, for a random sample of size $2$ with distribution function $F$ and density function $f$ I know that the joint density function of $X_{(j)}$ is given by $$f_{X_{(j)}}(t))=\frac{n!}{(j-1)!(n-j)!}\left[F(t)\right]^{j-1}\left[1-F(t)\right]^{n-j}f(t) \qquad -\infty

$$f_{X_{(j)}}(t)=\left\{\begin{array}{ll}\frac{1}{\sigma \sqrt{2\pi}}\left[1-\mathrm{erf}\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)\right]e^{-\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)^{2}} & \mbox{If }j=1 \\ \frac{1}{\sigma \sqrt{2\pi}}\left[1+\mathrm{erf}\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)\right]e^{-\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)^{2}} & \mbox{If }j=2 \end{array}\right. .$$ for $-\infty

Therefore, the expectation is

$$E(X_{(j)})=\left\{\begin{array}{ll}\frac{1}{\sigma \sqrt{2\pi}}{\displaystyle \int_{-\infty}^{\infty} t\left[1-\mathrm{erf}\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)\right]e^{-\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)^{2}}dt} & \mbox{If }j=1 \\ \frac{1}{\sigma \sqrt{2\pi}}{\displaystyle \int_{-\infty}^{\infty} t\left[1+\mathrm{erf}\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)\right]e^{-\left(\frac{t-\mu}{\sigma \sqrt{2}}\right)^{2}}dt} & \mbox{If }j=2 \end{array}\right. .$$

The problems begin when I want to calculate $\mathrm{Var}(X{(1)})$, $\mathrm{Var}(X{(2)})$ and $\mathrm{Cov}(X{(1)},X_{(2)})$ because I do not know the density function of the random variables of $X_{(j)}^{2}$ for $j=1,2$ and $X_{(1)}X_{(2)}$, I have not could calculate these densities, that is basically what I need, although I do not know if there is another way of making all this without having to calculate these densities.

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    Regarding the variances, note that for any random variable $X$ with PDF $f$, $$E(X^2)=\int_\mathbb Rx^2f(x)dx$$ Regarding the covariance, note that $$X_{(1)}X_{(2)}=X_1X_2$$ hence $$E(X_{(1)}X_{(2)})=E(X_1)E(X_2)=E(X_1)^2$$ Finally, I am surprised that your post does not mention that the joint PDF $g$ of $(X_{(1)},X_{(2)})$ for $(X_1,X_2)$ i.i.d. with PDF $f$ is simply $$g(x,y)=2f(x)f(y)\mathbf 1_{x – 2017-02-26
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    @Did I understand that $X_{(1)}X_{(2)}=X_{1}X_{2}$, hence $E(X_{(1)}X_{(2)})=E(X_{1}X_{2})$. Why do you say that $E(X_{(1)}X_{(2)})=E(X_1)E(X_2)$? Do you assume that $X_{1}$ and $X_{2}$ are iondependent? – 2017-02-26
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    Yes, this is a usual assumption when one mentions a random sample from a given distribution. Are you not making this independence assumption? – 2017-02-26

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