Discussion
Let X be a zero-mean unit variance Gaussian random variable. Define the random variable Y as :
\[ Y = \begin{cases} X & \text{if } |X| \le \alpha \\ -X & \text{if } |X| \gt \alpha \end{cases} \]
\(\alpha\) is some positive real number. What is the distribution of X + Y ?
Answer: C
X + Y is not Gaussian, with a distribution that is discontinuous at the origin.
Split on the event in the definition of Y:
- if \(|X| \le \alpha\), then \(Y = X\) and \(X + Y = 2X\);
- if \(|X| \gt \alpha\), then \(Y = -X\) and \(X + Y = 0\).
So X + Y equals 0 with probability \(P(|X| \gt \alpha) = 2(1 - \Phi(\alpha)) \gt 0\). A point mass at 0 makes the distribution function jump there, so it is discontinuous at the origin, and no Gaussian distribution has an atom. Away from 0, X + Y is spread over \([-2\alpha, 2\alpha]\) only, which is bounded.
The point of the question is that Y itself is N(0, 1), by the symmetry of X, yet X + Y is not Gaussian: two normal variables need not be jointly normal. Gaussian (0, 2) is what independence would give, and 'continuous distribution' misses the atom.
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