0
$\begingroup$

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).

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Thank you Dominik, so $\phi(v) = E[\exp(iv^t Y)] = i E[ \sum_{k = 0}^{\infty} k!^{-1} Y^k] = ?$ $\endgroup$
    – Monolite
    Dec 9, 2016 at 19:40
  • $\begingroup$ That formula is completely wrong in every way. $\endgroup$
    – Dominik
    Dec 9, 2016 at 19:42
  • $\begingroup$ I thought this was how the exponential of a vector was defined, should I try using Eulers formula? I apologize it's the first time going through these exercises. $\endgroup$
    – Monolite
    Dec 9, 2016 at 19:51
  • $\begingroup$ Could you please explain to me why it's wrong? I am here to learn, I am sorry if I make mistakes. $\endgroup$
    – Monolite
    Dec 9, 2016 at 20:08
  • $\begingroup$ Ok I think I understand my error now, so it should be: $\phi(v) = E[\exp(iv^t Y)] = E[\exp(i (v_1 + v_2) X)] = e^{-1/2 (v_1 + v_2)^2}$ so this means it is multivariate normal with mean $\mu = (0,0)$ and covariance matrix given by the 2 by 2 matrix of ones ? $\endgroup$
    – Monolite
    Dec 9, 2016 at 21:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .