Let K be a discrete random variable with PMF
pK(k)=⎧⎩⎨⎪⎪1/3,2/3,0if k=1,if k=2,otherwise.
Conditional on K=1 or 2, random variable Y is exponentially distributed with parameter 1 or 1/2, respectively.
Using Bayes' rule, find the conditional PMF pK∣Y(k∣y). Which of the following is the correct expression for pK∣Y(2∣y) when y≥0?
e^-y/2 / e^-y+e^-y/2 (Answered in full)
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!
To find the conditional PMF p(K|Y)(k|y), we can use Bayes' rule:
p(K|Y)(k|y) = [p(Y|K)(y|k) * pK(k)] / pY(y)
We are given that Y is exponentially distributed, so we have:
p(Y|K)(y|k) = λ * e^(-λy)
Where λ is the parameter of the exponential distribution for each value of k. In this case, λ = 1 for k = 1, and λ = 1/2 for k = 2.
Now, let's calculate the individual probabilities:
pK(1) = 1/3
p(Y|K)(y|1) = 1 * e^(-1y) = e^(-y)
pK(2) = 2/3
p(Y|K)(y|2) = (1/2) * e^(-1/2 * y) = e^(-y/2)
Since Y can take values 1 or 2, we have:
pY(y) = pY(y|1) * pK(1) + pY(y|2) * pK(2) = e^(-y) * (1/3) + e^(-y/2) * (2/3)
Now, let's substitute these values into Bayes' rule to find pK|Y(k|y):
p(K|Y)(k|y) = [p(Y|K)(y|k) * pK(k)] / pY(y)
For k = 2, we have:
p(K|Y)(2|y) = [e^(-y/2) * (2/3)] / [e^(-y) * (1/3) + e^(-y/2) * (2/3)]
Simplifying this expression gives us:
p(K|Y)(2|y) = e^(-y/2) / (e^(-y) + e^(-y/2))
Therefore, the correct expression for p(K|Y)(2|y) when y≥0 is:
p(K|Y)(2|y) = e^(-y/2) / (e^(-y) + e^(-y/2))
Hope this helps!
To find the conditional PMF pK∣Y(k∣y) using Bayes' rule, you need to use the following formula:
pK∣Y(k∣y) = (pY∣K(y∣k) * pK(k)) / pY(y)
In this case, you need to find pK∣Y(2∣y), where y ≥ 0. To do this, we need to calculate the numerator and denominator separately.
First, let's calculate the numerator. We know that Y follows an exponential distribution with parameter 1 or 1/2, given K = 1 or 2, respectively. Therefore, for K = 1, pY∣K(y∣1) = e^(-y), and for K = 2, pY∣K(y∣2) = (1/2)*e^(-y/2).
Now, let's calculate the denominator. We need to use the law of total probability to find pY(y):
pY(y) = ∑pY∣K(y∣k) * pK(k), for all possible values of k
In this case, we only have two possible values of k, which are k = 1 and k = 2. Therefore, the denominator becomes:
pY(y) = pY∣K(y∣1) * pK(1) + pY∣K(y∣2) * pK(2)
= e^(-y) * (1/3) + (1/2)*e^(-y/2) * (2/3)
= e^(-y)/3 + e^(-y/2)/3
Now, putting it all together, we can find pK∣Y(2∣y):
pK∣Y(2∣y) = (pY∣K(y∣2) * pK(2)) / pY(y)
= ((1/2)*e^(-y/2) * (2/3)) / (e^(-y)/3 + e^(-y/2)/3)
= (e^(-y/2) / 2) / (e^(-y) + e^(-y/2))
Therefore, the correct expression for pK∣Y(2∣y) when y ≥ 0 is e^(-y/2) / (e^(-y) + e^(-y/2)).