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: Concept Overview:
We aim to determine the probability that the minimum value of two independent and identically distributed (i.i.d.) discrete random variables equals a specific value. The key is to relate the event \(Z=k\) to events involving X and Y. When dealing with the minimum, using the survival function, \(P(Z>z)\), is a helpful approach.
Step 2: Core Formula:
Given \(Z = \min(X, Y)\) and the independence of X and Y:
\[ P(Z>z) = P(X>z \text{ and } Y>z) = P(X>z) P(Y>z) \]
For discrete variables, the probability mass function can be expressed via the survival function as:
\[ P(Z = k) = P(Z>k-1) - P(Z>k) \]
Or, equivalently, using the cumulative distribution function (CDF):
\[ P(Z = k) = P(Z \le k) - P(Z \le k-1) \]
Step 3: Step-by-Step Solution:
Our goal is to calculate \(P(Z=1)\). We will employ the survival function method.
\[ P(Z = 1) = P(Z>0) - P(Z>1) \]
Since X and 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, compute the 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, compute 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, compute \(P(Z=1)\):
\[ P(Z=1) = (1 - 2e^{-1} + e^{-2}) - (1 - 4e^{-1} + 4e^{-2}) \]
\[ P(Z=1) = 1 - 2e^{-1} + e^{-2} - 1 + 4e^{-1} - 4e^{-2} \]
\[ P(Z=1) = 2e^{-1} - 3e^{-2} \]
To match the options, express 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 Result:
The probability \(P(Z=1)\) is \( \frac{2e-3}{e^2} \).