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Given that $X$ and $Y$ are jointly standard normally distributed, find the $Prob(X\cdot 1_{\{Y\leq c\}} \leq z)$. Where $1_{\{Y\leq c\}}$ is an indicator variable. c is a constant.

$$P(X\cdot 1_{\{Y\leq c\}} \leq z)=P(X\cdot 1\leq z,Y\leq c) + P(X\cdot 0\leq z,Y\geq c) =P(X\cdot 1\leq z,Y\leq c)\\ $$

Am I on the right path?

Kofi
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1 Answers1

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Assuming by "jointly standard normally distributed" you mean that $\mathrm{Cov}(X,Y)=0$, we can compute this probability by conditioning on $Y$: \begin{align} \mathbb P(X\cdot\mathsf 1_{\{Y\leqslant c\}}\leqslant z) &= \mathbb P(X\cdot\mathsf 1_{\{Y\leqslant c\}}\leqslant z\mid Y\leqslant c)\mathbb P(Y\leqslant c)\\ &= \mathbb P(X\leqslant z\mid Y\leqslant c)\mathbb P(Y\leqslant c)\\ &= \mathbb P(X\leqslant z)\mathbb P(Y\leqslant c)\\ &= \Phi(z)\Phi(c), \end{align} where $\Phi$ denotes the distribution function of a standard normal random variable.

Math1000
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  • Thanks for the response. Would it be possible to explain intuitively the first line of your solution -- $\mathbb{P}(X\cdot 1 {{Y\leq c}} \leq z)= \mathbb{P}(X\cdot1{{Y⩽c}}⩽z∣Y⩽c)\mathbb{P}(Y⩽c)$? – Kofi Feb 21 '20 at 16:20
  • That is just the definition of conditional probability... – Math1000 Feb 21 '20 at 19:47