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

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 $$H$$.

• What have you tried? – Arthur Aug 4 '19 at 15:44
• The question has hint which is condition on x. I am confused how to calculate CDF on H – dodka Aug 4 '19 at 15:53

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 $$1$$.