Find the expectation of $Z=min\{X,Y\}$ which $X~N(\mu,{\sigma}^2)$,$Y~N(\mu,{\sigma}^2)$

Question

Find the expectation of $Z=min\{X,Y\}$ which $X\sim N(\mu,{\sigma}^2)$,$Y\sim N(\mu,{\sigma}^2)$. $X$ and $Y$ are independent random variables.

This is how far I go: According to order statistics, I know that $PDF$ of $Z$ is:$$f_Z(z)=2(1-F(z))f(z)$$

where $f(z)$ is the $PDF$ of $X$ or $Y$ and $F(z)$ is the anti-derivative(CDF) of $f(z)=\frac{1}{\sqrt{2\pi}\sigma}$exp$\left(-\frac{(z-\mu)^2}{2\sigma^2}\right)$

So $$E(Z)=\int_{-\infty}^{\infty}zf_Z(z)dz=\int_{-\infty}^{\infty}2(1-F(z))f(z)dz=2E(Z)-2\int_{-\infty}^{\infty}zf(z)F(z)dz=2\mu-2A$$

I stuck in solving $A$, any help in solving $A$ or the original problem would be appreciated. Thanks

$\color{red}{Edit}$ [By following Did's comment]

Let $Z=\mu+\sigma Z_0$ where $Z_0$ coresponds to the standard normal distribution. Then my original $$E(Z)=u+min\{Z_1,Z_2\}=\mu+\sigma(2\mu-2A)=\mu+0-2\sigma A$$ $$A=\int_{-\infty}^{\infty}z\varphi(z)\Phi(z)dz=\int_{-\infty}^{\infty}\varphi^2(z)dz=\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{-z^2}dz=\frac{1}{2\pi}\sqrt{\pi}=\frac{1}{2\sqrt{\pi}}$$ So $\color{red}{E(Z)=\mu-2\sigma A=\mu-\frac{\sigma}{\sqrt\pi}}$

Answer 1

We exploit the formula

$$\min \{ X,Y \}=X-\min \{ X-Y,0 \}.$$

Hence $E[\min \{ X,Y \}]=E[X]-E[\min \{ X-Y,0 \}]=\mu-E[\min \{ X-Y,0 \}]$.

It remains to compute the second term. Since $X,Y$ are iid $N(\mu,\sigma^2)$, $X-Y$ is $N(0,2 \sigma^2)$. So the second term is

$$\int_{-\infty}^0 0 dz + \int_0^\infty z n_{0,2\sigma^2}(z) dz$$

where $n_{m,s^2}$ represents the pdf of a $N(m,s^2)$ variable. This last integral can be explicitly calculated to be $\frac{\sigma}{\sqrt{\pi}}$. So the overall answer is $\mu-\frac{\sigma}{\sqrt{\pi}}$.

This approach was suggested in Mark Joshi's answer but the OP expressed that they didn't follow that answer, so I filled in some details.

Answer 2

you can view the whole as an option pricing problem. $$ \min(X,Y) = X - (X-Y)_+ $$ So what is $$ E((X-Y)_+) $$ ?

$X-Y$ is distributed $N(0,2\sigma^2).$ Write it as $X-Y = \sigma \sqrt{2} Z$ with $Z$ standard normal, then $$ E((X-Y)_+) = \sqrt{2} \sigma E(Z_+). $$ Now $$ E(Z_+) = \frac{1}{\sqrt{2\pi}} \int \limits_{0}^{\infty} z e^{-z^2/2}dz = \frac{1}{\sqrt{2\pi}} . $$ Done.

For more info look up the Bachelier model. (see eg my book More Mathematical Finance.)