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\frac{y-2}5\right)=F_X\left(\frac{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(\frac{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+\frac{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)`