Question

The random variable Q is uniform on [0, 1]. Conditioned on Q = q, the random variable X is Bernoulli with parameter q.

Answer 1

In this scenario, you have a two-level random variable system:

1. The first random variable \( Q \) is uniformly distributed over the interval \([0, 1]\). This means the probability density function (pdf) of \( Q \) is given by:
\[
f_Q(q) =
\begin{cases}
1 & \text{if } 0 \leq q \leq 1 \\
0 & \text{otherwise}
\end{cases}
\]

2. Given a realization of \( Q = q \), the second random variable \( X \) follows a Bernoulli distribution with parameter \( q \). The Bernoulli distribution takes on the value 1 with probability \( q \) and 0 with probability \( 1 - q \). Therefore, the conditional probability mass function of \( X \) given \( Q = q \) is:
\[
P(X = 1 \mid Q = q) = q, \quad P(X = 0 \mid Q = q) = 1 - q.
\]

To analyze this setup further, we can derive the marginal distribution of \( X \) by integrating over the possible values of \( q \). The joint distribution of \( X \) and \( Q \) can be specified as follows:

\[
P(X = x, Q = q) = P(X = x \mid Q = q) f_Q(q).
\]

### For \( X = 1 \):
\[
P(X = 1, Q = q) = P(X = 1 \mid Q = q) f_Q(q) = q \cdot 1 = q, \quad \text{for } 0 \leq q \leq 1.
\]

To find the marginal probability \( P(X = 1) \):
\[
P(X = 1) = \int_0^1 P(X = 1, Q = q) dq = \int_0^1 q \, dq = \left[\frac{q^2}{2}\right]_{0}^{1} = \frac{1}{2}.
\]

### For \( X = 0 \):
\[
P(X = 0, Q = q) = P(X = 0 \mid Q = q) f_Q(q) = (1 - q) \cdot 1 = 1 - q, \quad \text{for } 0 \leq q \leq 1.
\]

To find the marginal probability \( P(X = 0) \):
\[
P(X = 0) = \int_0^1 P(X = 0, Q = q) dq = \int_0^1 (1 - q) \, dq = \left[q - \frac{q^2}{2}\right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2}.
\]

### Summary:
- Marginal probability of \( X = 1 \):
\[
P(X = 1) = \frac{1}{2}
\]

- Marginal probability of \( X = 0 \):
\[
P(X = 0) = \frac{1}{2}
\]

Thus, \( X \) is uniformly distributed between 0 and 1 as well, with the probability of it being 1 or 0 both equal to 1/2.