If $B$ is a random variable following a known distribution (in my case $B$ follows a Pareto distribution with parameters $\alpha = 1.5$ and $\lambda = 1000$) and $R$ is independent of $B$ following a normal distribution of mean $0.08$ and standard deviation $0.2$. Is there any way to find the cdf of $X=B \times 1_{\lbrace R \leq -0.1 \rbrace}$. I tried a lot (by conditionning on the variable $R$) and it did not lead me to somewhere. Any hints?

  • $\begingroup$ The distribution of $1_{\{R\,\le\,-1\}}$ is Bernoulli with expected value about $ 0.1586553$. No information about the distribution of $R$ beyond that is relevant here. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 2 '15 at 0:29

I'm assuming that's a simple product (and not, say, a cartesian product).

Let $\{A = 0\} = \{R\le -0.1\}$, and $\{A = 1\} = \{R > -0.1 \}$.

Let $P(A = 1) = p$

\begin{align} P(X \le x) &= \sum_{i \in \{0,1\}} P(X \le x ~|A = i)P(A = i) \\ &= P(0 \le x) (1-p) + P(B \le x)p \\ &= (1-p) u(x) + P(B \le x)p \end{align}

where $u(x)$ is the step function. You can play with the value of $u(0)$ to make the CDF left or right continuous, whatever your convention may be.

  • $\begingroup$ Thanks for your answer! So, from what I ve read, you're just using the total probability law. So independency is irrelevant here? $\endgroup$ – mich95 Oct 2 '15 at 0:38
  • $\begingroup$ Well, in this case, it seems so. If it was say, addition instead, it would matter a lot more. Think of how you'd calculate the distribution of the sum of two independent gaussians. $\endgroup$ – stochasticboy321 Oct 2 '15 at 0:52

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