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
-
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.
-
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: \[ f_Q(q | K = 1) = 2q \quad \text{for } 0 \leq q \leq 1. \]
1 answer