# Standard normal probability distribution

I ran into this problem.

If $ξ$ is a random variable with a standard normal distribution and η is a random variable such that: $η = \left\lbrace\right. ξ$, if $\left\|ξ\right\| \le 2$ $, -ξ$ if $\left\|ξ\right\| \gt 2$

Compute a distribution of the number variable $η$.

I am not sure whether I understand what they want me to compute. Isn't the distribution of η just a density function $φ(x) = \left(\frac{e^{-x^2/2}}{\sqrt {2\pi}}\right)$ for $x \le 2$ and $-φ(x)$ for $x \gt 2$?

• Notice that if you have $-φ(x)$ then your probabilities will be negative, which cannot be the case – WaveX Dec 11 '16 at 19:10
• so the distribution of η is $φ(x)$ for $x \ge 2$ and $0$ otherwise? – mathew7k5b Dec 11 '16 at 19:13
• η is $φ(x)$ for $x≤2$ and $x≥ -2$ because of the condition $∥ξ∥≤2$ – WaveX Dec 11 '16 at 19:17
• Hint: eta is standard normal. – Did Dec 11 '16 at 19:24
• @Did how can it be proved? I think I still don't see what happens with $η$ when $\left\|ξ\right\| \gt 2$ – mathew7k5b Dec 11 '16 at 19:31