Mathematics: 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)= ?

Let x and y be independent continuous random variables

jeanine239

30.12.2021

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 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

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

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

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

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

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breemills9953

22.07.2022

N is to the left of the zero which means it is a negative number.

P is the the right of zero which means it is a positive number.

A negative number is less than a positive number, so the answer would be:

The number with the least value is N

The number with the greatest value is P

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