Exponential distribution with variable mean

Question

I am studying for an exam coming tomorrow and there's one exercise i couldn't really solve:

In a region there are two types of rains: short and long ones. 70% of them are short. The time T (in hours) of the duration of a rain follows an exponential distribution with mean 1 when the rain is short, and mean 4 when the rain is long. The variance of T (in hours^2) is: (Answer is 7.39)

Don't really know where to start so any hints are truly appreciated.

I know that the variance is defined as $V(x)= E(x^2)-(E(x))^2$ where in a single exponential distribution $V(x) = 2/\lambda^2 - 1/\lambda^2 = 1/\lambda^2$

Answer 1

Ok i think i solved the exercise, sorry for any notation problems in my solution:

$ E(T)=P(X_{short\ rain})*E(T|X_{short\ rain}) + P(X_{long\ rain})*E(T|X_{long\ rain})$

$E(T)=0.7*\frac{1}{\lambda_{short\ rain}}+0.3*\frac{1}{\lambda_{long\ rain}}=0.7*\frac{1}{\frac{1}{1}}+0.3*\frac{1}{\frac{1}{4}}=1.19$

$E(T^2)=P(X_{short\ rain})*E(T^2|X_{short\ rain}) + P(X_{long\ rain})*E(T^2|X_{long\ rain})$

$E(T^2)=0.7*\frac{2}{\lambda_{short\ rain}^2}+0.3*\frac{2}{\lambda_{long\ rain}^2}=0.7*\frac{2}{(\frac{1}{1})^2}+0.3*\frac{1}{(\frac{1}{4})^2}=11$

$Var(T) = E(T^2) - E(T)^2 = 11 -1.19^2 = 7.39$

Answer 2

You have that $T|\lambda \sim exp(1/\lambda)$ and $\lambda$ follows a discrete distribution with values 1 with probability 0.7 and 4 with probability 0.3. You can then use the iterated variance formula given by $$ Var(T) = E[Var(T|\lambda)]+Var(E[T|\lambda]). $$ Note that $Var(T|\lambda) = \lambda^2$ and $E[T|\lambda] = \lambda$. Then $$ \begin{aligned} Var(T) &= E[\lambda^2]+Var(\lambda) \\ &= E[\lambda^2]+E[\lambda^2]-E[\lambda]^2\\ &= 2E[\lambda^2]-E[\lambda]^2. \end{aligned} $$ From here, you can calculate explicitly these values through the distribution of $\lambda$.