What is the probability of a random variable less than a value?

Question

If $X$ and $Y$ are independent random variable following the normal distribution $N(\mu, \sigma^2)$ with mean $\mu$ and variance $\sigma^2$ such that $X \sim N(\mu_{X}, \sigma_{X}^2)$ and $Y \sim N(\mu_{Y}, \sigma_{Y}^2)$, and if $T$ is some constant, then I want to find the probability $P(X-Y<T)$.

According to this solution, I have done the following:

First, I define a new random variable $Z =X-Y$.

Then, $$\mathbb{E}(Z) = \mathbb{E}(X-Y) = \mathbb{E}(X) - \mathbb{E}(Y) = \mu_{X}-\mu_{Y}=\mu_{Z}.$$

$$\mathbb{V}(Z) = \mathbb{V}(X-Y) = \mathbb{V}(X) + (-1)^2 \mathbb{V}(Y) = \sigma_{X}^2 + \sigma_{Y}^2 = \sigma_{Z}^2.$$

Thus, I obtain the probability:

$$\begin{equation} \begin{aligned} \mathbb{P}(X-Y<T) &= \mathbb{P}(Z<T) \\[6pt] &= \mathbb{P} \Bigg( \frac{Z+\mu_{Z}}{\sqrt{\sigma_{Z}^2}} < \frac{T+\mu_{Z}}{\sqrt{\sigma_{Z}^2}} \Bigg) \\[6pt] &= \Phi \Bigg( \frac{T+\mu_{Z}}{\sqrt{\sigma_{Z}^2}} \Bigg) \\[6pt] \\[6pt] \end{aligned} \end{equation}$$

I have two questions:

1. Could someone explain how $\mathbb{P} \Bigg( \frac{Z+\mu_{Z}}{\sqrt{\sigma_{Z}^2}} < \frac{T+\mu_{Z}}{\sqrt{\sigma_{Z}^2}} \Bigg)$ came into existence? Or direct me to a relevant material that I could read upon.

2. And of course, is the derivation correct?

Answer 1

Actually, it should be: $$P(Z<T)=P\left(\frac{Z-\mu_Z}{\sigma_Z}<\frac{T-\mu_Z}{\sigma_Z}\right)=\Phi\left(\frac{T-\mu_Z}{\sigma_Z}\right)$$ In the other post, the mean is $-1$ so I guess it seemed like a summation to you. This method of subtracting from the mean and dividing by deviation is applied only to convert $Z$ into a standard normal RV (by making it zero mean and unit variance) so that we can use Z-table available to calculate the probability.