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

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.

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{aligned} \mathbb{P}(X-Y

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?

Actually, it should be: $$P(Z In the other post, the mean is $$-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.
• My bad. In my handwritten notes, I subtracted the mean of $X$ and $Y$. Thank you very much for the correction though. Regarding the conversion to standard normal RV in the other post, it looks like it is converted to the generalized normal RV. Mar 18, 2020 at 14:39
• You're welcome. $Z$ is a normal RV and $(Z-\mu_z)/\sigma_Z$ is standard normal RV. So, $Z$ is converted into standard normal to use the tables available. For the generalized normal, it's already in that family: en.wikipedia.org/wiki/Generalized_normal_distribution And, $\Phi(t)$ is the CDF of standard normal distribution. Mar 18, 2020 at 14:43