Question

Let K be a discrete random variable with PMF

P K (k) =1/4, if k = 1,1/2 if k2, 1/4, if k = 3, 0 otherwise.

Conditional on K = 1, 2 or 3, random variable Y is exponentially distributed with parameter 1 1/2 or 1/3, respectively

Using Bayes rule, find the conditional PMF p K|Y ( k |y) Which of the following is the correct expression for p K cap Y ( 2 |y) when y >= 0 ?

Answer 1

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 \( K \):

- For \( K = 1 \): \( Y \sim \text{Exponential}(1) \)
- Probability density function: \( P(Y=y | K=1) = e^{-y} \) for \( y \ge 0 \)

- 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 \( y \ge 0 \)

- 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:

1. For \( K=1 \):
\[
P(Y=y | K=1) P(K=1) = e^{-y} \cdot \frac{1}{4} = \frac{1}{4} e^{-y}
\]

2. For \( K=2 \):
\[
P(Y=y | K=2) P(K=2) = 2e^{-2y} \cdot \frac{1}{2} = e^{-2y}
\]

3. 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:

1. For \( k = 1 \):
\[
P(K=1 | Y=y) = \frac{P(Y=y | K=1) P(K=1)}{P(Y=y)} = \frac{\frac{1}{4} e^{-y}}{P(Y=y)}
\]

2. For \( k = 2 \):
\[
P(K=2 | Y=y) = \frac{P(Y=y | K=2) P(K=2)}{P(Y=y)} = \frac{e^{-2y}}{P(Y=y)}
\]

3. 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 \( K \).