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Let K be a discrete random variable with PMF

PK (k)=
1/4, if k 1, 1/2, if k 2, 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 YCY ( k |y) Which of the following is the correct expression for p KY ( 2 |y) when y >= 0

1 answer

To find the conditional PMF \( p_{K|Y}(k | y) \) using Bayes' rule, we first need to utilize the relationship:

\[ p_{K|Y}(k | y) = \frac{p_{Y|K}(y | k) \cdot P_K(k)}{p_Y(y)} \]

Step 1: Find \( p_Y(y) \)

The probability density function (PDF) of \( Y \) can be found by using the law of total probability:

\[ p_Y(y) = \sum_{k} p_{Y|K}(y | k) \cdot P_K(k) \]

The distributions for \( Y \) based on the values of \( K \) are as follows:

  • For \( K = 1 \) (with \( P_K(1) = \frac{1}{4} \)), \( Y \sim \text{Exponential}(1) \):

    \[ p_{Y|K}(y | 1) = e^{-y} \quad (y \geq 0) \]

  • For \( K = 2 \) (with \( P_K(2) = \frac{1}{2} \)), \( Y \sim \text{Exponential}\left(\frac{1}{2}\right) \):

    \[ p_{Y|K}(y | 2) = \frac{1}{2} e^{-\frac{y}{2}} \quad (y \geq 0) \]

  • For \( K = 3 \) (with \( P_K(3) = \frac{1}{4} \)), \( Y \sim \text{Exponential}\left(\frac{1}{3}\right) \):

    \[ p_{Y|K}(y | 3) = \frac{1}{3} e^{-\frac{y}{3}} \quad (y \geq 0) \]

Now compute \( p_Y(y) \):

\[ p_Y(y) = p_{Y|K}(y | 1) \cdot P_K(1) + p_{Y|K}(y | 2) \cdot P_K(2) + p_{Y|K}(y | 3) \cdot P_K(3) \]

\[ = e^{-y} \cdot \frac{1}{4} + \frac{1}{2} e^{-\frac{y}{2}} \cdot \frac{1}{2} + \frac{1}{3} e^{-\frac{y}{3}} \cdot \frac{1}{4} \]

\[ = \frac{1}{4} e^{-y} + \frac{1}{4} e^{-\frac{y}{2}} + \frac{1}{12} e^{-\frac{y}{3}} \]

Step 2: Find \( p_{Y|K}(y | k) \cdot P_K(k) \)

We will compute this for \( k = 1, 2, 3 \):

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

  2. For \( K = 2 \): \[ p_{Y|K}(y | 2) \cdot P_K(2) = \frac{1}{2} e^{-\frac{y}{2}} \cdot \frac{1}{2} = \frac{1}{4} e^{-\frac{y}{2}} \]

  3. For \( K = 3 \): \[ p_{Y|K}(y | 3) \cdot P_K(3) = \frac{1}{3} e^{-\frac{y}{3}} \cdot \frac{1}{4} = \frac{1}{12} e^{-\frac{y}{3}} \]

Step 3: Combine and find \( p_{K|Y}(k | y) \)

Now substitute all these into Bayes' rule for each value of \( k \):

For \( K = 1 \):

\[ p_{K|Y}(1 | y) = \frac{p_{Y|K}(y | 1) \cdot P_K(1)}{p_Y(y)} = \frac{\frac{1}{4} e^{-y}}{\frac{1}{4} e^{-y} + \frac{1}{4} e^{-\frac{y}{2}} + \frac{1}{12} e^{-\frac{y}{3}}} \]

For \( K = 2 \):

\[ p_{K|Y}(2 | y) = \frac{p_{Y|K}(y | 2) \cdot P_K(2)}{p_Y(y)} = \frac{\frac{1}{4} e^{-\frac{y}{2}}}{\frac{1}{4} e^{-y} + \frac{1}{4} e^{-\frac{y}{2}} + \frac{1}{12} e^{-\frac{y}{3}}} \]

For \( K = 3 \):

\[ p_{K|Y}(3 | y) = \frac{p_{Y|K}(y | 3) \cdot P_K(3)}{p_Y(y)} = \frac{\frac{1}{12} e^{-\frac{y}{3}}}{\frac{1}{4} e^{-y} + \frac{1}{4} e^{-\frac{y}{2}} + \frac{1}{12} e^{-\frac{y}{3}}} \]

Summary of Conditional PMF for \( K \)

The desired expression for \( p_{K|Y}(2 | y) \) is thus:

\[ p_{K|Y}(2 | y) = \frac{\frac{1}{4} e^{-\frac{y}{2}}}{\frac{1}{4} e^{-y} + \frac{1}{4} e^{-\frac{y}{2}} + \frac{1}{12} e^{-\frac{y}{3}}} \]

This matches the correct form of the conditional PMF \( p_{K|Y}(2 | y) \) for \( y \geq 0 \).