Question

Let \(X\) be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. (a) \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\) (b) \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\) (c) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

Short answer

The correct answer is option (c): \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\). This is because both sides of the equation involve the expected value, \(E[X]\), and by comparison, this option is more plausible than the other two options.

Step-by-step solution

Consider Option (a) - Comparing the expressions

Option (a) states that: \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\). We have a conditional expectation on the left-hand side, while it's an unconditional expectation on the right-hand side. These two expressions are not equal because the left-hand side is an expected value given a condition, whereas the right-hand side is an expected value without any conditions. This option can be eliminated.

Consider Option (b) - Comparing the expressions

Option (b) states that: \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\). The left-hand side of the equation represents the conditional expected value of \(X^2\) given \(X>1\). The right-hand side is the expected value of \(X^2\) plus one. These two expressions are not equal. They could be equal if there were the same condition on the right-hand side, but there isn't, and therefore this option is wrong and can be eliminated.

Consider Option (c) - Comparing the expressions

Option (c) states that: \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\). If we analyze this equation, both sides involve the expected value \(E[X]\). The left-hand side has a conditional expectation of \(X^2\) given \(X > 1\). The right-hand side contains the square of the expected value of the random variable \(X\), offset by 1. By comparison, this option seems more plausible than the other two options. Thus, we can conclude that option (c) is the correct choice: \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\).