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Let $X$ and $Y$ be independent continuous random variables that are uniformly distributed on $(0,1)$. Let $H:=(X+2)Y$.

Find the probability $\textsf{P}(\ln H \geq z)$, where $z$ is a given number that satisfies $e^z<2$.

The answer should be a function of $z$.


The question has hint, which is "condition on $x$". I am confused how to calculate CDF on $H$.

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    $\begingroup$ What have you tried? $\endgroup$ – Arthur Aug 4 '19 at 15:44
  • $\begingroup$ The question has hint which is condition on x. I am confused how to calculate CDF on H $\endgroup$ – dodka Aug 4 '19 at 15:53
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For any given $z$ that satisfies the condition, $\textsf{P}(\,\ln H \geq z) = \textsf{P}( H \geq e^z)= \textsf{P}(\, (X+2)Y\geq e^z)$ such that the this desired probability is the area (since the distribution is uniform) within the unit square $\{x,y \} \in [0, 1]^2$ above the contour level curve $$ (x+2)y \geq e^z \implies y \geq \frac{ e^z }{ x+2 }$$ which is the familiar hyperbola $\frac1x$ shifted by two and scaled by the given constant $e^z$ (this is where the condition $e^z < 2$ comes from).

By the area "above" the curve, it means the usual integration at a given $x$ (which is equivalent to conditioning on $x$ here) from $\frac{ e^z }{ x+2 }$ up to $1$.

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