2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.