Let $(X,Z)$ be bivariate normal with parameters
$\mu_X := E(X) = 1, \mu_Z := E(Z ) = 1, \sigma_X^2 := Var(X) = 1$,
$ \sigma_Z^2
:= Var(Z ) = 1$,
and the correlation coefficient of (X, Z ) is $\rho$ with density
$$f(x,z)=\dfrac{1}{2\pi\sqrt{1-\rho^2}}\exp(-\dfrac{1}{2(1-\rho^2)}((x-1)^2-2\rho (x-1)(z-1)+(z-1)^2)).$$
Let $Y=\min(X,Z)$. Then what is $E[Y]$
For the above what I have so far is
$\textbf{MY Attempt}$
we first transform the variables so that $X=1+X_1$ and $Z=1+X_2$. Then
\begin{align*}
\mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 – \mathbb{P}\left\{ X_1 > y, X_2 > y \right\} \\
&= 1 – \int_{y}^{\infty} \int_y^{\infty} f_{X_1,X_2}(s,t)dsdt \\
&= 1 -\int_y^{\infty} f_{X_2}(t)\int_y^{\infty} f_{X_1|X_2}(s,t) ds dt
\end{align*}
where
$$
f_{X_2} (t)=\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}, \qquad f_{X_1|X_2}(s,t)=\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(s-\rho t)^2}{2(1-\rho^2)}},
$$
and so
\begin{align*}
\mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 – \int_{y}^{\infty} \varphi(t) \left(1 – \Phi\left( \frac{y -\rho t}{\sqrt{1-\rho^2}}\right) \right) dt. \\
&= 1 – \int_{y}^{\infty} \varphi(t) \Phi\left( \frac{\rho t – y }{\sqrt{1-\rho^2}} \right) dt
\end{align*}
To get the density we differentiate with respect to $y$ giving
\begin{align*}
f_{Y}(y) &= -\frac{\partial}{\partial y}\int_{y}^{\infty}\varphi(t)\Phi\left(\frac{\rho t – y}{\sqrt{1-\rho^2}}\right) dt \\
&= \varphi(y)\Phi\left(\frac{\rho y – y}{\sqrt{1-\rho^2}}\right) + \int_{y}^{\infty}\varphi(t) \frac{1}{\sqrt{1-\rho^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\rho t – y)^2}{2(1-\rho^2)}}dt \tag{1}
\end{align*}
Completing the square of the last term in $(1)$ we have
\begin{align*}
\varphi(t)\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(\rho t-y)^2}{2(1-\rho^2)}} &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left((1-\rho^2)t^2 +(\rho t -y)^2 \right)} \\
&=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left(t^2 – 2 t \rho y + \rho^2 y^2 + (1-\rho^2)y^2 \right)} \\
&= \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}.
\end{align*}
So putting this back in to $(1)$ we get
\begin{align*}
f_{Y}(y) &= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\int_y^{\infty} \frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}dt \\
&= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\left( 1 – \Phi\left(\frac{y – \rho y}{\sqrt{1-\rho^2}}\right)\right) \\
&= 2\varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right).
\end{align*}
$\textbf{Hence I think I have the density of random variable Y. From here its difficult for me to find the expectation of Y.}$
Best Answer
You can use the following $$\min(X,Z) = \frac{(X+Z)}{2}-\frac{|X-Z|}{2} $$
the distribution of $W =(X -Z)$ is normal with mean $0$ and variance $2(1-\rho)$ Now this is a random variable and we can take out the distribution of $|W|$ as $$\mathbb{P}(|W|\leq t) = \mathbb{P}(W\leq t)-\mathbb{P}( W \leq -t)=F_W(t) - F_W(-t)$$ Hence we will have $$f_{|W|}(t) = f_{W}(t)+f_{W}(-t) = 2f_{W}(t) $$ there we can get the $\mathbb{E}(|W|) = \frac{2}{\sigma\sqrt{2\pi}}\displaystyle\int_0^{\infty}z\exp{ \frac{-z^2}{2\sigma^2} }dz = \sigma\sqrt{\frac{2}{\pi}}$
Hence we will have the answer using $\mathbb{E}\Big[\min(X,Z)\Big] = .5\mathbb{E}\Big[X\Big]+.5\mathbb{E}\Big[Z\Big]-.5\mathbb{E}\Big[|W|\Big]$.