5.1.5 problem 6

a. For $X\sim \mathrm{Unif}(0,1)$, $F\left(k\right)=k$ for $0<k<1$, $E\left(X\right)=1∕2$, $V\mathrm{ar}\left(X\right)=1∕12$, $\mathrm{STD}\left(X\right)=1/2\sqrt{3}$.
Then the probability X is within one standard deviation of its mean is $P(\frac{\sqrt{3}-1}{2\sqrt{3}}<X<\frac{\sqrt{3}+1}{2\sqrt{3}})=F\left(\frac{\sqrt{3}+1}{2\sqrt{3}}\right)-F\left(\frac{\sqrt{3}-1}{2\sqrt{3}}\right)=\frac{1}{\sqrt{3}}$.

The probability that X is within two standard deviations of its mean is 1, as the mean plus two standard deviations $1/2+1/\sqrt{3}$ exceeds 1 and the mean minus two standard deviations is less than 0 - since X always takes values between 0 and 1, X is always within 2 standard deviations of its mean. Similarly, it is always within 3 standard deviations of the mean.

b. We have $E\left(X\right)=1$ and $V\mathrm{ar}\left(X\right)=1$. $F\left(k\right)=1-e^{-k}$. Also note that $P(X<0)=0$ for an exponential distribution.

1 standard deviation: $P(0<X<2)=F\left(2\right)-F\left(0\right)=F\left(2\right)=1-e^{-2}$
2 standard deviation: $P(-1<X<3)=P(0<X<3)=F\left(3\right)-F\left(0\right)=F\left(3\right)=1-e^{-3}$
3 standard deviation: $P(-2<X<4)=P(0<X<4)=F\left(4\right)-F\left(0\right)=F\left(4\right)=1-e^{-4}$

c. If $Y\sim \mathrm{Expo}(1∕2)$, then $Y=2X$ where $X\sim \mathrm{Expo}\left(1\right)$, $E\left(Y\right)=2$, $V\mathrm{ar}\left(Y\right)=4$, $\mathrm{STD}\left(Y\right)=4$. In general, we note that if $Y\sim \mathrm{Expo}\left(\lambda \right)$ then $Y=X∕\lambda$ and $E\left(Y\right)=1∕\lambda$ and $\mathrm{STD}\left(Y\right)=1∕\lambda$.

Then we can realize the following pattern: the probability that Y is $n$ standard deviations away from its mean is $P(-(n-1)/\lambda <Y<(n+1\left)\lambda \right)=P(0<Y<(n+1)/\lambda )=F\left(\right(n+1)/\lambda )=1-e^{\lambda (n+1)∕\lambda }=1-e^{n+1}$