Question

Let X and Y be independent continuous random variables that are uniformly distributed on (0,1). Let H=(X+2)Y. Find the probability P(lnH≥z) where z is a given number that satisfies ez<2. Your answer should be a function of z.Hint: Condition on X.P(lnH≥z1)= ? Let X be a standard normal random variable, and let FX(x) be its CDF. Consider the random variable Z=FX(X). Find the PDF fZ(z) of Z. Note that fZ(z) takes values in (0,1) .fZ(z)= ?

Answer 1

Looks like there are two distinct problems here.

• "Let X and Y be independent ..."

$P(\ln H\ge z)=P(\ln((X+2)Y)\ge z)=P((X+2)Y\ge e^z)=P\left(Y\ge\dfrac{e^z}{X+2}\right)$

Conditioning on $X$ involves interpreting this probability for all possible fixed values of $X=x$:

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_xP\left(Y\ge\dfrac{e^z}{X+2}\text{ and }X=x\right)$

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_xP\left(Y\ge\dfrac{e^z}{X+2}\mid X=x\right)P(X=x)$

We have $x\in(0,1)$, and over this domain $P(X=x)=1$. Since $X=x$ is fixed, we can omit the conditioning notation to leave us with

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1P\left(Y\ge\dfrac{e^z}{x+2}\right)\,\mathrm dx$

$Y$ has CDF

$P(Y\le y)=F_Y(y)=\begin{cases}0&\text{for }y1\end{cases}$

Then we can write our integral as

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1\left(1-F_Y\left(\frac{e^z}{x+2}\right)\right)\,\mathrm dx$

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1\left(1-\frac{e^z}{x+2}\right)\,\mathrm dx$

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=x-e^z\ln(x+2)\bigg|_{x=0}^{x=1}$

$P\left(Y\ge\dfrac{e^z}{X+2}\right)=\boxed{1-e^z(\ln2-\ln3)}$

• "Let X be a standard normal ..."

Using the method of distribution functions, we have

$F_Z(z)=P(Z\le z)=P(F_X(X)\le z)=P(X\le {F_X}^{-1}(z))=F_X({F_X}^{-1}(z))=z$

$\implies\boxed{f_Z(z)=\begin{cases}1&\text{for }z\in(0,1)\\0&\text{otherwise}\end{cases}}$