This was a question in my probability theory class recently:
There are $5$ balls, numbered $1-5$, in a bag. We take $3$ of those
balls at random. What is the expected value of the sum of the $3$
balls taken out of the bag?
Here's one approach that my instructor suggeseted:
Let $I_i$ be the Indicator random variable for ball with the number $i$, which is 1 if the ball is selected and $0$ otherwise.
We then calculate the the required expected value of the sum as :
$$
\sum_{i=1}^{5} iE[I_i]=\frac{3}{5}(1 + 2 + 3 + 4 + 5)\boxed{=9}
$$
I don't understand why this approach works. Why would the summation of the multiplication of the expectation of the indicator random variable with the number that it is associated with giving me the correct answer?
Best Answer
The indicator $I_i$ is a Bernoulli random variable that takes the value $1$ if ball $i$ was one of the 3 balls you chose and $0$ otherwise. Recall the expected value of a Bernoulli random variable is the probability of success, which in this case is $3/5$.
Notice the sum of the values of the 3 balls you chose can be expressed as $$S=\sum_{i=1}^5 iI_i.$$
Use linearity of expectation to compute $E[S]$ and you are done.