Suppose that U ∼ U [0, 1] is a uniform random variable with range [0, 1] (i.e., U has probability density function (PDF) fU (p) = 1 for all p ∈ [0, 1] and fU(p) = 0 for all p ̸∈ [0,1]). Suppose furthermore that X ∈ {1,2,...} is a discrete random variable such that the conditional probability mass function (PMF) of X given U = p ∈ [0, 1] is geometric with success probability p. in other words, for all p ∈ [0, 1], the conditional probability mass function (PMF) of X given U = p is
( (1−p)n−1p ifn∈{1,2,...}, pX|U=p(n) = 0 otherwise.
(i) Specify P (X = 1) and P (X = 2). Explain.

To solve this problem, we need to compute the probabilities \( P(X = 1) \) and \( P(X = 2) \) by using the given information. 

### Given Information:

- \( U \sim U[0, 1] \) is a uniform random variable with the PDF:
  \[
  f_U(p) = 
  \begin{cases} 
  1 & \text{if } 0 \leq p \leq 1, \\
  0 & \text{otherwise.}
  \end{cases}
  \]

- \( X \) is a discrete random variable, and the conditional PMF of \( X \) given \( U = p \) is geometric with success probability \( p \). Specifically:
  \[
  P(X = n \mid U = p) = p(1-p)^{n-1}, \quad n = 1, 2, 3, \dots
  \]

### To Find: \( P(X = 1) \) and \( P(X = 2) \)

#### **Step 1: Compute \( P(X = 1) \)**

The probability \( P(X = 1) \) can be computed by marginalizing over the uniform random variable \( U \):
\[
P(X = 1) = \int_{0}^{1} P(X = 1 \mid U = p) f_U(p) \, dp
\]

Substitute the given conditional probability and the PDF of \( U \):
\[
P(X = 1) = \int_{0}^{1} p \cdot 1 \, dp = \int_{0}^{1} p \, dp
\]

Evaluate the integral:
\[
P(X = 1) = \left[\frac{p^2}{2}\right]_{0}^{1} = \frac{1}{2}
\]

#### **Step 2: Compute \( P(X = 2) \)**

Similarly, \( P(X = 2) \) can be computed as:
\[
P(X = 2) = \int_{0}^{1} P(X = 2 \mid U = p) f_U(p) \, dp
\]

Substitute the given conditional probability:
\[
P(X = 2) = \int_{0}^{1} p(1-p) \cdot 1 \, dp = \int_{0}^{1} p(1-p) \, dp
\]

Expand and evaluate the integral:
\[
P(X = 2) = \int_{0}^{1} (p - p^2) \, dp = \left[\frac{p^2}{2} - \frac{p^3}{3}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}
\]

### **Conclusion:**
- \( P(X = 1) = \frac{1}{2} \)
- \( P(X = 2) = \frac{1}{6} \)

These probabilities result from integrating the conditional probability over the uniform distribution of \( U \).

Would you like more details, or do you have any other questions?

Here are 5 related questions:
1. How would you compute \( P(X = 3) \)?
2. What is the expected value \( \mathbb{E}[X] \)?
3. How does the distribution of \( U \) affect \( P(X = n) \) for general \( n \)?
4. How would the result change if \( U \) were uniform on a different interval?
5. What would be the effect of changing the geometric distribution's success probability?

**Tip:** Integrals involving polynomials can often be simplified by recognizing patterns or standard integral formulas.

Explore how to calculate conditional probabilities with a uniform random variable and a geometrically distributed discrete random variable. This problem involves integrating conditional probability mass functions and applying concepts from probability theory and integral calculus.

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