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Expectation and Linearity (1+1+1 = 3 marks)

Linearity of expectation is often a powerful tool when other combinatorial (pure counting-based) approaches prove cumbersome.

Consider a problem where \(n\) students' pens are randomly and uniformly shuffled and given back to the students. This implies that any particular student gets back a pen that is uniformly distributed among all the pens. We want to find the expected number of students who got back their own pens.

1. Denote by \(A_k\) the event that the \(k\)th student got back his/her pen, and by \(X_k\) its indicator random variable. What is the expected value of \(X_k\)?

2. Denote by \(X\) the number of students who got back their own pens. Find a relation between \(X\) and \(X_1, \ldots, X_n\).

3. Use the relation you found above, along with the linearity of expectation, to find \(E[X]\).

Answer :

the expected value of Xk is 1/n, and the expected value of X, the number of students who got back their own pens, is 1. Linearity of expectation allows us to find the expected value of X by taking the sum of the expected values of the indicator random variables Xk.

In this problem, we are asked to find the expected number of students who got back their own pens. To do this, we can use the concept of linearity of expectation.

First, let's define some terms. Let Ak be the event that the kth student got back his/her pen, and let Xk be its indicator random variable. The indicator random variable Xk takes on the value of 1 if the kth student got back his/her pen, and 0 otherwise.

Now, we want to find the expected value of Xk. The expected value of Xk is the probability that the event Ak occurs. In this case, since the pens are randomly and uniformly shuffled, the probability that the kth student gets back his/her own pen is 1/n, where n is the total number of students. Therefore, the expected value of Xk is 1/n.

Next, let's define X as the number of students who got back their own pens. We can express X as the sum of the indicator random variables X1, X2, ..., Xn. In other words, X = X1 + X2 + ... + Xn.

Using linearity of expectation, we can find the expected value of X by taking the sum of the expected values of X1, X2, ..., Xn. Since the expected value of each Xk is 1/n, we have:

E[X] = E[X1 + X2 + ... + Xn] = E[X1] + E[X2] + ... + E[Xn] = 1/n + 1/n + ... + 1/n = n * (1/n) = 1.

Therefore, the expected number of students who got back their own pens is 1.
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