Find the integral of $P(XLet X,Y be r.v following a normal distribution ($\mu_X,\mu_Y$ and $\sigma_X^2,\sigma_Y^2 >0$) that are independent. Let $z:=X-Y$. What integral shows $P(X<Y)$. Do not compute it
I can not find the integral but I am fairly sure on how to compute it. What is the integral? Is it just the integral of the normal distribution from $\pm \infty$? I know the linear property applies for $Z$.
 A: Simply calculate $P(Z<0)$
$$Z\sim N(\mu_X-\mu_Y;\sigma_X^2+\sigma_Y^2)$$
To calculate it you need the tables but you can express it in terms of $\Phi$

The text gives you an important hint: find the distribution of $Z=X-Y$
As per mean and variance properties:
$$E(aX+bY)=aE(X)+bE(Y)$$
$$V(aX+bY)=a^2V(X)+b^2V(Y)$$
and for gaussian's properties $Z$ has the distribution I showed you thus
$$\mathbb{P}[Z<0]=\int_{-\infty}^0 f_Z(z)dz=\Phi\left(\frac{\mu_Y-\mu_X}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)$$
A: I think it is useful for you a numerical example:
Let $X\sim N(1;1)$ and $Y\sim N(2;1)$ and let's $X,Y$ independent.
We want to calculate $P(X<Y)=P(X-Y<0)$
To do that we first derive $f_Z(z)$.
As per what I told you in my previous post, $Z\sim N(-1;2)$
Thus
$$P(Z<0)=\Phi\left(\frac{0-(-1)}{\sqrt{2}}\right)=\Phi\left(\frac{1}{\sqrt{2}}\right)=76.025\%$$
A: Ok: Now that I know your background it is easier for me to help you...
Observe that your exercise explicitly states

What integral shows P(X<Y)? Do not compute it

thus  what they want from you is this
$$\mathbb{P}[X<Y]=\int_{-\infty}^{0}\frac{1}{\sqrt{2\pi(\sigma_X^2+\sigma_Y^2)}}e^{-[z-(\mu_X-\mu_Y)]^2/[2(\sigma_X^2+\sigma_Y^2)]} dz$$
...and nothing else.
