Oh, Bayes' rule, huh? Sounds like a mathematical puzzle to me. Let's give it a shot!
According to the information given, we have that pK(k=1) = 1/3, pK(k=2) = 2/3, and the conditional distribution of Y given K=1 is exponential with parameter 1, while the conditional distribution of Y given K=2 is exponential with parameter 1/2.
Now, we want to find pK∣Y(2∣y), the conditional probability that K=2 given that Y=y. Using Bayes' rule, we have:
pK∣Y(2∣y) = pY∣K(y∣2) * pK(2) / pY(y)
Since we're looking for an expression for pK∣Y(2∣y) for y≥0, we don't have to worry about the denominator pY(y). We just need to focus on the other terms.
Given that K=2, the conditional distribution of Y is exponential with parameter 1/2. So we have:
pY∣K(y∣2) = (1/2)e^(-1/2)y
And pK(2) = 2/3, as stated earlier.
Now, let's put it all together: pK∣Y(2∣y) = (1/2)e^(-1/2)y * (2/3).
So the correct expression for pK∣Y(2∣y) when y≥0 is indeed e^-y/2 / e^-y + e^-y/2.
Hope that gives you a little mathematical entertainment!