Question

Let \( X \) be a random variable with pdf \[ f(x) = \begin{cases} e^{-x}, & x > 0 \\ 0, & x \le 0 \end{cases} \] Define \( Y = \lfloor X \rfloor \), the greatest integer less than or equal to \( X \). Then \(E(Y^2)\) is equal to:

• A. \(\frac{e(e+1)}{e-1}\)
• B. \(\frac{e+1}{(e-1)^2}\)
• C. \(\frac{(e+1)^2}{e-1}\)
• D. \(\frac{(e+1)^2}{(e-1)^2}\)

Hint:

Whenever the random variable is an integer part of a continuous exponential variable, convert its pmf and use geometric series formulas for expectations.

Answer 1

The Correct Option is B

Step 1: Express the pmf of \(Y\).
For integer \( k \ge 0 \), \[ P(Y = k) = P(k \le X<k+1) = e^{-k} - e^{-(k+1)} = e^{-k}(1 - e^{-1}) \]
Step 2: Compute \(E(Y^2)\).
\[ E(Y^2) = \sum_{k=0}^{\infty} k^2 P(Y = k) = (1 - e^{-1}) \sum_{k=0}^{\infty} k^2 e^{-k} \]
Step 3: Use the known series formula.
\[ \sum_{k=0}^{\infty} k^2 r^k = \frac{r(1 + r)}{(1 - r)^3}, \quad |r|<1 \] Substitute \(r = e^{-1}\): \[ E(Y^2) = (1 - e^{-1}) \frac{e^{-1}(1 + e^{-1})}{(1 - e^{-1})^3} = \frac{(e+1)}{(e-1)^2} \] Final Answer: \[ \boxed{\frac{e+1}{(e-1)^2}} \]