Question

Let, X and Y be independent and identically distributed Poisson(1) variables. If, Z = min(X, Y) then, P(Z = 1) is:

• A. \( \frac{e-3}{e^2} \)
• B. \( \frac{2e-3}{e^2} \)
• C. \( \frac{2e-3}{2e^2} \)
• D. \( \frac{1-2e}{e^2} \)

Hint:

For problems involving the minimum or maximum of independent random variables, using the CDF (\(F_Z(z)\)) or the survival function (\(S_Z(z) = P(Z>z)\)) is often much more efficient than considering all possible combinations of outcomes. For min, \(S_{\min}(z) = S_X(z)S_Y(z)\). For max, \(F_{\max}(z) = F_X(z)F_Y(z)\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the probability of the minimum of two i.i.d. discrete random variables being equal to a specific value. The key is to express the event \(Z=k\) in terms of events involving X and Y. A useful technique for the minimum is to work with the survival function, \(P(Z>z)\).

Step 2: Key Formula or Approach:
For \(Z = \min(X, Y)\) with X and Y being independent: \[ P(Z>z) = P(X>z \text{ and } Y>z) = P(X>z) P(Y>z) \] For discrete variables, the probability mass function can be found using the survival function: \[ P(Z = k) = P(Z>k-1) - P(Z>k) \] Or, equivalently, using the CDF: \[ P(Z = k) = P(Z \le k) - P(Z \le k-1) \]
Step 3: Detailed Explanation:
We want to find \(P(Z=1)\). Let's use the survival function approach. \[ P(Z = 1) = P(Z>0) - P(Z>1) \] Since X, Y are i.i.d., \(P(Z>z) = [P(X>z)]^2\). First, we need the probabilities for a single Poisson(1) variable, X. The PMF is \(P(X=k) = \frac{e^{-1}1^k}{k!} = \frac{e^{-1}}{k!}\). - \(P(X=0) = e^{-1}\) - \(P(X=1) = e^{-1}\) Now, calculate the required survival probabilities for X: - \(P(X>0) = 1 - P(X=0) = 1 - e^{-1}\) - \(P(X>1) = 1 - P(X \le 1) = 1 - (P(X=0) + P(X=1)) = 1 - (e^{-1} + e^{-1}) = 1 - 2e^{-1}\) Now, calculate the survival probabilities for Z: - \(P(Z>0) = [P(X>0)]^2 = (1 - e^{-1})^2 = 1 - 2e^{-1} + e^{-2}\) - \(P(Z>1) = [P(X>1)]^2 = (1 - 2e^{-1})^2 = 1 - 4e^{-1} + 4e^{-2}\) Finally, calculate \(P(Z=1)\): \[ P(Z=1) = (1 - 2e^{-1} + e^{-2}) - (1 - 4e^{-1} + 4e^{-2}) \] \[ P(Z=1) = 1 - 2e^{-1} + e^{-2} - 1 + 4e^{-1} - 4e^{-2} \] \[ P(Z=1) = 2e^{-1} - 3e^{-2} \] To match the options, we can write this with a common denominator of \(e^2\): \[ P(Z=1) = \frac{2}{e} - \frac{3}{e^2} = \frac{2e}{e^2} - \frac{3}{e^2} = \frac{2e-3}{e^2} \]
Step 4: Final Answer:
The probability P(Z=1) is \( \frac{2e-3}{e^2} \).