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I randomly chose either Alice or Bob to go catch a penguin for me with equal probabilities of chosing either person. Let $I=0$ if I chose Alice and $I=1$ if I chose Bob. Alice can catch a penguin in time $T_1 \sim Exponential(\lambda_1)$. Bob can catch a penguin in time $T_2 \sim Exponential(\lambda_2)$. Let $T$ be the time it takes for a penguin to be caught. What is the variance of $T$?

I started this problem with:

$$ Var(T) = E(Var(T|I)) + Var(E(T|I)) $$

However, I'm not sure of how to calculate either $E(Var(T|I))$ or $Var(E(T|I))$.

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With regard to the question itself, $\text{var}(T\mid I)$ is a random variable that takes on two values depending on whether $I$ is $0$ or $1$. What is its average value? Similarly, $E[T\mid I]$ is a random variable that takes on two values. What is its average? What is its variance? You will get a lot farther if you write down the two values mentioned above (in the two cases) explicitly and then proceed, instead of trying to deal with mystical magical formulas in generality. – Dilip Sarwate Dec 13 '12 at 21:00
up vote 4 down vote accepted

HINT: Start by computing $E(T|I)$ and $Var(T|I)$, these are just the expectation and variance of an exponentialy distributed variable thus

$$E(T|I)=\frac{1}{\lambda_{I+1}} \text{ and } Var(T|I)=\frac{1}{(\lambda_{I+1})^2} \; .$$

Can you take it from here?

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