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.