Question

The random variable K is geometric with a parameter which is itself a uniform random variable Q on [0, 1] Find the value f Q given K(0.5|1) of the conditional PDF of Q given that K =1 Hint: Use the result in the last segment

Answer 1

To solve for the conditional probability density function \( f_Q(q | K = 1) \) of the random variable \( Q \) given that \( K = 1 \), where \( K \) is a geometric random variable with parameter \( Q \) (which itself is uniformly distributed on \([0, 1]\)), we can apply the law of total probability and Bayes' theorem.

### Step 1: Understand the distributions involved

1. **The random variable \( Q \)** is uniformly distributed over \([0, 1]\):
\[
f_Q(q) = 1 \quad \text{for } 0 \leq q \leq 1
\]
and \( f_Q(q) = 0 \) otherwise.

2. **The random variable \( K \)** given \( Q = q \) follows a geometric distribution with success parameter \( q \):
\[
P(K = k | Q = q) = (1 - q)^{k - 1} q
\]
where \( k \) is a positive integer. In this case, we are interested in the case when \( K = 1 \):
\[
P(K = 1 | Q = q) = q
\]

### Step 2: Compute the marginal distribution \( P(K = 1) \)

We can obtain the marginal distribution of \( K \) using the law of total probability, integrating over all possible values of \( Q \):
\[
P(K = 1) = \int_0^1 P(K = 1 | Q = q) f_Q(q) \, dq = \int_0^1 q \cdot 1 \, dq
\]
Calculating this integral:
\[
P(K = 1) = \int_0^1 q \, dq = \left[ \frac{q^2}{2} \right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2}
\]

### Step 3: Apply Bayes' theorem

Now we can apply Bayes' theorem to find \( f_Q(q | K = 1) \):
\[
f_Q(q | K = 1) = \frac{P(K = 1 | Q = q) f_Q(q)}{P(K = 1)}
\]
Substituting the values:
- \( P(K = 1 | Q = q) = q \)
- \( f_Q(q) = 1 \)

Thus:
\[
f_Q(q | K = 1) = \frac{q \cdot 1}{\frac{1}{2}} = 2q \quad \text{for } 0 \leq q \leq 1
\]

### Conclusion

The final result for the conditional PDF of \( Q \) given \( K = 1 \) is:
\[
f_Q(q | K = 1) = 2q \quad \text{for } 0 \leq q \leq 1.
\]