Given a sequence $\{ X_{n} : n \in \mathbb{N} \}$ of standard normally distributed random variables, what is the probability that $X_{1}+X_{2}<1$?

Question

I'm looking for a general formula that can be used to compute the probability that a sum of standard normal random variables is above a certain constant $a$. So for example the probability that $X_{1}+X_{2}+X_{3}+X_{4}<3$. I've tried to play around with the distribution of a sum of standard normal distributed random variables but wasn't able to come up with a real formula. Any help is appreciated.

Answer 1

For a R.V. $Y \sim N(0,\sigma^2)$, by definition:

$P(Y \leq y) = \int_{-\infty}^y \frac{1}{\sigma \sqrt{2 \pi}}e^{-x^2/2 \sigma^2}dx$.

Let's substitute $u = x/\sigma$ to get:

$P(Y \leq y) = \int_{-\infty}^{y/\sigma} \frac{1}{\sqrt{2 \pi}} e^{-u^2/2} du = \Phi \left(\frac{y}{\sigma} \right)$

where $\Phi$ is the cumulative distribution function of a standard normal variable (which has zero mean and unit variance). You can evaluate $\Phi$ in any mathematical package or from printed or online Z tables.

In your specific case, as many have pointed out, $ Y = \sum_{i=1}^n X_i$ has mean 0 and variance $\sigma^2 = n$, which means you need to evaluate $\Phi(y/n)$ to compute $P(Y \leq y)$.