Any help would be appreciated. I am kind of losing my mind over this problem.
Consider
$T(\gamma)=\int_0^\infty\Phi(-\beta(x+\gamma))\Phi'(z+x-\gamma)\, dx$.
Consider, the CDF function defined from T as follow:
$F(y)=\frac{1}{T(\gamma)}\int_0^y\Phi(-\beta(x+\gamma))\Phi'(z+x-\gamma)\,dx$.
How can we show $F$'s variance is increasing in $\gamma$.
Note that $\Phi(x)$ is the CDF of standard normal distribution.
Is this a right kind of argument: (How can I formalize it)
We can rewrite $T$ as follow:
$T=\frac{\beta}{2\pi}\int_{0}^\infty\int_{-y}^0e^{-\frac{1}{2}(x+v-z)^2-\frac{1}{2}(y+v)^2\beta^2}\, dx\, dy$.
When we increase $v$, we decrease means of unconditional bivariate distribution (X,Y). Now if we condition on $-Y<X<0$ and $Y>0$, we cut the distribution along the same place. So the variance should increase?
or this one? We can say
$T=\frac{1}{2\pi}\int_{-1}^\infty\int_{2}^\infty\beta \gamma^2 e^{-\frac{1}{2}(\gamma x+z)^2-\frac{1}{2}(\gamma x+\gamma y)^2\beta^2}\, dy \, dx$ The inside of the integral is PDF of bivariate normal distribution for $(\gamma X,\gamma(X+Y))$. When $\gamma$ is larger, the variance of $(\gamma X,\gamma(X+Y))$ is larger (I do not even know what does variance of two random variable mean). Hence, if we condition it on $X>-1$ and $Y>2$, the variance should still be larger.