This is an exam question given at a statistic and probability course;
If $X \sim N_{0,1}$ and $Y = \left( \begin{array}{ccc} X \\ X \end{array} \right)$.
1) find a Borellian $H \subset R^2$ of Lebesgue measure zero s.t. $P(Y \in H) = 1$; deduce from this that $Y$ can't be absolutely continuous.
2) Calculate the characteristic function of Y.
3) Using the characteristic function show that Y has a multi-normal distribution.
My attempt at a solution:
1) choose $H = R \times \{ 0 \}$ then $P( Y \in H) = P( \left( \begin{array}{ccc} X^{-1}(R) \\ X^{-1}(R) \end{array} \right)) = 1 $ but the Lebesgue measure of $H \in R^2$ is zero. This means that Y is not absolutely continuous since if it was I could write its distribution as a Lebesgue integral on $R^2$ but the lebesgue integral on a set of measure zero is always zero so I have my contradiction.
2) For point two I am stuck since without a density how can I compute the expected value?
Could anyone help me understand how to proceed? (and if what I have done up until now is right).
Part 1 is incorrect. For your choice of $H$, you have $P(Y \in H) = P(X \in R, X = 0) = 0$. Instead, you should take $H = \{(x, x) \mid x \in \mathbb R\}$.
For part 2, just remember the definition of the characteristic function $\phi(v) = E[\exp(iv^t Y)]$ and use the fact that $X \sim \mathcal{N}(0, 1)$. Part 3 then follows from part 2.