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 $x \gt 2$?

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

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