To solve the problem, we need to compute the conditional probability mass function \( P(K=k | Y=y) \) using Bayes' theorem.
Bayes' theorem states:
\[ P(K=k | Y=y) = \frac{P(Y=y | K=k) P(K=k)}{P(Y=y)} \]
We will consider the cases for \( k = 1, 2, 3 \).
Step 1: Find \( P(Y=y | K=k) \)
Given that \( Y \) is exponentially distributed with different parameters based on the value of \( K \):
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For \( K = 1 \): \( Y \sim \text{Exponential}(1) \)
- Probability density function: \( P(Y=y | K=1) = e^{-y} \) for \( y \ge 0 \)
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For \( K = 2 \): \( Y \sim \text{Exponential}(1/2) \)
- Probability density function: \( P(Y=y | K=2) = \frac{1}{(1/2)} e^{-y/(1/2)} = 2e^{-2y} \) for \( y \ge 0 \)
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For \( K = 3 \): \( Y \sim \text{Exponential}(1/3) \)
- Probability density function: \( P(Y=y | K=3) = \frac{1}{(1/3)} e^{-y/(1/3)} = 3e^{-3y} \) for \( y \ge 0 \)
Step 2: Find \( P(K=k) \)
The PMF of \( K \) is given as:
\[ P(K=1) = \frac{1}{4}, \quad P(K=2) = \frac{1}{2}, \quad P(K=3) = \frac{1}{4} \]
Step 3: Find \( P(Y=y) \)
Using the law of total probability:
\[ P(Y=y) = \sum_{k=1}^{3} P(Y=y | K=k) P(K=k) \]
Calculating each component:
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For \( K=1 \): \[ P(Y=y | K=1) P(K=1) = e^{-y} \cdot \frac{1}{4} = \frac{1}{4} e^{-y} \]
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For \( K=2 \): \[ P(Y=y | K=2) P(K=2) = 2e^{-2y} \cdot \frac{1}{2} = e^{-2y} \]
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For \( K=3 \): \[ P(Y=y | K=3) P(K=3) = 3e^{-3y} \cdot \frac{1}{4} = \frac{3}{4} e^{-3y} \]
Combining these gives:
\[ P(Y=y) = \frac{1}{4} e^{-y} + e^{-2y} + \frac{3}{4} e^{-3y} \]
Step 4: Compute \( P(K=k | Y=y) \)
Now we can use Bayes' theorem:
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For \( k = 1 \): \[ P(K=1 | Y=y) = \frac{P(Y=y | K=1) P(K=1)}{P(Y=y)} = \frac{\frac{1}{4} e^{-y}}{P(Y=y)} \]
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For \( k = 2 \): \[ P(K=2 | Y=y) = \frac{P(Y=y | K=2) P(K=2)}{P(Y=y)} = \frac{e^{-2y}}{P(Y=y)} \]
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For \( k = 3 \): \[ P(K=3 | Y=y) = \frac{P(Y=y | K=3) P(K=3)}{P(Y=y)} = \frac{\frac{3}{4} e^{-3y}}{P(Y=y)} \]
Answer
Thus, the conditional PMF is given by:
\[ P(K=k | Y=y) = \frac{P(Y=y | K=k) P(K=k)}{P(Y=y)}, \quad k = 1, 2, 3. \]
For example, the expression \( P(K=2 | Y=y) \) is:
\[ P(K=2 | Y=y) = \frac{e^{-2y}}{\frac{1}{4} e^{-y} + e^{-2y} + \frac{3}{4} e^{-3y}} \]
For \( y \ge 0 \). Thus, \( P(K \cap Y = 2 | Y=y) \) simplifies to the expression found above for \( P(K=2 | Y=y) \).
In summary, the complete conditional PMF \( P(K=k | Y=y) \) is fully developed based on the analysis of the distribution of \( Y|K \) and the PMF of \( K \).