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What would you do to find probability when a uniform random variable is the parameter for the distribution of another uniform random variable.

ie: $Z \sim Unif(0,1)$ $Y \sim Unif(0,Z)$

And we are interested in finding the probability of $Y$ either $>, <$, or $=$.

I feel as if I have to use a double integral but am not sure where to go.

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1 Answer 1

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Guide:

Use conditional probability, first figure out $P(Y \le y|Z=z)$ and $f_Z(z)$

Now you can compute the following:

\begin{align} P(Y \le y) &= \int_y^1 P(Y \le y|Z=z)f_Z(z)\, dz \end{align}

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  • $\begingroup$ Could you complete the computation? I think I'm making a mistake in my own calculations and would like to verify the result of the integral. $\endgroup$ Commented May 3, 2020 at 3:55
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    $\begingroup$ I was hoping that the OP would post his working after following the guide. Any particular step/term that you are uncertain? $\endgroup$ Commented May 3, 2020 at 4:00
  • $\begingroup$ The result of the integral diverges, leading me to believe the marginal of $y$ may not exist. $\endgroup$ Commented May 3, 2020 at 4:07
  • $\begingroup$ I made a mistake earlier, $z$ can only take values that are larger than $y$. Thanks for pointing out the mistake. $\endgroup$ Commented May 3, 2020 at 4:11
  • $\begingroup$ I am having a lot of trouble coming up with what would 𝑃(𝑌≤𝑦|𝑍=𝑧) be since Y and Z are dependent random variables. I know that 𝑓𝑍(𝑧) = 1 but am not sure what to do $\endgroup$
    – pluto
    Commented May 3, 2020 at 19:32

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