The random variable K is geometric with a parameter which is itself a uniform random variable Q on [0,1]. Find the value fQ|K(0.5|1) of the conditional PDF of Q, given that K=1. Hint: Use the result in the last segment.
the answer is : 1
To find the conditional PDF of Q, given that K=1, we need to find the value fQ|K(0.5|1).
The first step is to find the joint PDF fQ,K(q,k). Since K is a geometric random variable with parameter Q, we know that the PDF of K is given by fK(k) = (1-q)^(k-1)*q, where q is the parameter of the geometric distribution.
Next, we need to find the marginal PDF of Q, which is denoted as fQ(q). The marginal PDF of a random variable can be found by integrating the joint PDF over all possible values of the other random variables. In this case, the possible values of K are 1, 2, 3, and so on.
Since K=1, we can substitute k=1 into the joint PDF and integrate over all possible values of Q (from 0 to 1) to find the marginal PDF of Q.
So, the marginal PDF can be computed as:
fQ(q) = ∫ fQ,K(q,1) dK = ∫ fQ,K(q,1) * fK(1) dK = ∫ fQ,K(q,1) * (1-q)^(1-1)*q dK = q * fQ,K(q,1).
Now, we know that the marginal PDF fQ(q) is equal to q * fQ,K(q,1). To find fQ|K(0.5|1), the conditional PDF of Q given that K=1, we need to divide the joint PDF fQ,K(q,1) by the marginal PDF fQ(q).
Therefore, fQ|K(0.5|1) = fQ,K(0.5,1) / fQ(0.5) = q * fQ,K(0.5,1) / (q * fQ(0.5)).
The hint suggests using the result from the last segment, which implies that the joint distribution is given by fQ,K(q,k) = fK(k|q) * fQ(q), where fK(k|q) is the conditional PDF of K given Q=q.
In this case, since K is geometric with parameter Q, the conditional PDF of K given Q=q is simply given by fK(k|q) = (1-q)^(k-1)*q.
Therefore, we can substitute fK(k|q) = (1-q)^(k-1)*q and fQ(q) = q into the equation for fQ|K(0.5|1) to get:
fQ|K(0.5|1) = (1-q)^(k-1)*q * q / (q * q * fQ(0.5)) = (1-q)^(k-1) / fQ(0.5).
Now, we just need to compute fQ(0.5) and substitute it into the equation to get the final answer for fQ|K(0.5|1).