Suppose X is an exponential random variable with density function fX(x) = e−x, x > 0.
1
Also, the conditional density of Y given X is Poisson with mean X; i.e. xn −x
P(Y =n|X =x)= n! e , n∈{0,1,2,...}. Find P(Y = 1).
Also, find the generating function E tY and hence find the expectation of Y , confirming your answer with a direct calculation of the mean viaE(Y)=E[E(Y |X)].

We are given the following setup:

- \( X \) is an exponential random variable with probability density function (pdf) \( f_X(x) = e^{-x} \), for \( x > 0 \).
- \( Y \) is conditionally Poisson given \( X = x \), with mean \( X = x \). Thus, \( P(Y = n | X = x) = \frac{x^n e^{-x}}{n!} \), for \( n \in \{0, 1, 2, \dots \} \).

### Part 1: Find \( P(Y = 1) \)

We are tasked with finding the unconditional probability \( P(Y = 1) \). To compute this, we use the law of total probability by integrating over the conditional distribution of \( Y = 1 \) given \( X = x \):

\[
P(Y = 1) = \int_0^\infty P(Y = 1 | X = x) f_X(x) \, dx
\]

The conditional probability \( P(Y = 1 | X = x) \) is given by:

\[
P(Y = 1 | X = x) = \frac{x^1 e^{-x}}{1!} = x e^{-x}
\]

Thus, the expression becomes:

\[
P(Y = 1) = \int_0^\infty x e^{-x} \cdot e^{-x} \, dx = \int_0^\infty x e^{-2x} \, dx
\]

This integral can be solved using standard integration techniques. The integral of \( x e^{-ax} \) (where \( a = 2 \)) is known to be:

\[
\int_0^\infty x e^{-2x} \, dx = \frac{1}{4}
\]

Thus, we have:

\[
P(Y = 1) = \frac{1}{4}
\]

### Part 2: Find the generating function \( E[t^Y] \)

Next, we find the generating function of \( Y \), which is \( E[t^Y] \). This is the expected value of \( t^Y \), which can be written as:

\[
E[t^Y] = E[E[t^Y | X]]
\]

Given that \( Y | X = x \) is Poisson with mean \( x \), the moment generating function of a Poisson random variable is \( e^{x(t - 1)} \). Therefore:

\[
E[t^Y | X = x] = e^{x(t - 1)}
\]

Thus, we have:

\[
E[t^Y] = \int_0^\infty e^{x(t - 1)} f_X(x) \, dx = \int_0^\infty e^{x(t - 1)} e^{-x} \, dx
\]

Simplifying the exponent:

\[
E[t^Y] = \int_0^\infty e^{x(t - 2)} \, dx
\]

The integral of \( e^{ax} \) (for \( a < 0 \)) is \( \frac{1}{-a} \). Here, \( a = t - 2 \), so:

\[
E[t^Y] = \frac{1}{2 - t}, \quad \text{for} \ t < 2
\]

### Part 3: Find \( E[Y] \) from the generating function

To find \( E[Y] \), we differentiate the generating function with respect to \( t \) and then evaluate at \( t = 1 \):

\[
E[Y] = \frac{d}{dt} \left( \frac{1}{2 - t} \right) \Bigg|_{t=1}
\]

Differentiating \( \frac{1}{2 - t} \):

\[
\frac{d}{dt} \left( \frac{1}{2 - t} \right) = \frac{1}{(2 - t)^2}
\]

Evaluating at \( t = 1 \):

\[
E[Y] = \frac{1}{(2 - 1)^2} = 1
\]

### Part 4: Confirm \( E[Y] \) via \( E[E(Y | X)] \)

We can confirm this result by calculating \( E[Y] \) directly using the law of total expectation:

\[
E[Y] = E[E(Y | X)]
\]

Since \( Y | X = x \) is Poisson with mean \( x \), we know that \( E(Y | X = x) = x \). Therefore:

\[
E[Y] = E[X]
\]

Since \( X \) is exponentially distributed with mean \( 1 \) (for the given pdf \( f_X(x) = e^{-x} \)):

\[
E[X] = 1
\]

Thus, \( E[Y] = 1 \), confirming the result obtained from the generating function.

### Conclusion:

- \( P(Y = 1) = \frac{1}{4} \)
- \( E[Y] = 1 \)

Would you like more details or clarifications on any part of the solution?

Here are some follow-up questions you might explore:

1. What is the probability \( P(Y = 2) \) under the same conditions?
2. How would the expectation of \( Y \) change if \( X \) had a different distribution?
3. What is the variance of \( Y \)?
4. How does the moment generating function \( E[e^{tY}] \) relate to the generating function \( E[t^Y] \)?
5. Can you derive a general formula for \( P(Y = n) \) for any \( n \)?

**Tip**: When working with moment generating functions (MGFs), differentiating them gives valuable insights into the moments (e.g., mean, variance) of a random variable.

Suppose X is an exponential random variable with density function fX(x) = e−x, x > 0. Also, the conditional density of Y given X is Poisson with mean X; i.e. P(Y =n|X =x) = (x^n * e−x) / n!, n ∈ {0,1,2,...}. Find P(Y = 1). Also, find the generating function E[t^Y] and hence find the expectation of Y, confirming your answer with a direct calculation of the mean via E(Y) = E[E(Y | X)].

This problem explores the relationship between an exponential random variable X and a Poisson-distributed Y given X, using probability theory. Learn how to compute P(Y=1), find the generating function E[t^Y], and calculate the expected value of Y through the law of total expectation. Suitable for advanced probability and statistics students.

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