Question

g let X be a normally distributed random variable with mean 3 and variance 4. a) Let Y = 5X+2. What is the distribution of Y? What are its mean and variance? b) Find P(Y<10). Find P(X<10). c) What is the 99th percentile of the distribution of Y? d) What is the 99th percentile of the distribution of X? e) What is the distribution of W = exp(Y)? What are its mean and variance?

Answer 1

a. Let $F_X(x)$ be the CDF of $X$. The CDF of $Y$ is

$F_Y(y)=P(Y\le y)=P(5X+2\le y)=P\left(X\le\dfrac{y-2}5\right)=F_X\left(\dfrac{y-2}5\right)$

which is to say, $Y$ is also normally distributed, but with different parameters. In particular,

$E[Y]=E[5X+2]=5E[X]+2=17$

$\mathrm{Var}[Y]=\mathrm{Var}[5X+2]=5^2\mathrm{Var}[X]=100$

b. Using the appropriate CDFs, we have

$P(Y<10)=F_Y(10)=F_X\left(\dfrac{10-2}5\right)=F_X(1.6)\approx0.242$

$P(X<10)=F_X(10)\approx0.9998$

c. The 99th percentile for any distribution $D$ is the value of $d_{0.99}$ such that $P(D\le d_{0.99})=0.99$, i.e. all values of $d$ below $d_{0.99}$ make up the lower 99% of the distribution.

We have

$P(Y\le y_{0.99})=0.99\implies y_{0.99}\approx40.26$

d. On the other hand, the 99th percentile for $X$ is

$P(X\le x_{0.99})=0.99\implies x_{0.99}\approx7.653$

e. We have

$F_W(w)=P(W\le w)=P\left(e^Y\le w\right)=P(Y\le\ln w)=F_Y(\ln w)$

which suggests that $\ln W$ is normally distributed, or $W$ is log-normally distributed. Recall that the moment-generating function for $Y$ is

$M_Y(t)=\exp\left(17t+\dfrac{100t^2}2\right)$

But we also have

$M_Y(t)=E[e^{tY}]=E[e^{t\ln W}]=E[W^t]$

Then

$E[W]=M_Y(1)=e^{67}$

and

$E[W^2]=M_Y(2)=e^{234}\implies\mathrm{Var}[W]=E[W^2]-E[W]^2=e^{234}-e^{134}$