We are considering the expected loss if we take decision d6. The first case is if ∅=1/4 in actual. And we are considering how much expected loss we will suffer if we take a1(∅=1/4), a2(∅=1/2) and a3(∅=3/4). But in d6, our decision is: 1/2 for x=0 and 3/4 for x=1. It means we are not taking a1(∅=1/4) in this decision function. So, its prob. will be zero. Thats why P(a1|∅=1/4) = 0.
Thank you very much for your reply. But, how is the probability P(a2|∅1=1/4) = 3/4 and P(a3|∅1=1/4) = 1/4 calculated?
P(a2|∅1=1/4) is the prob. of a2(our decision is ∅=1/2) given that actual ∅=1/4. And we will decide ∅=1/2 when x=0 if you see it in table in part (ii). So, the prob. is x=0 for Bin(1,1/4) = ¹C0 ×(1/4)^0×(3/4)¹ that is 3/4. Similarly, for P(a3|∅=1/4), a3 means we are deciding it 3/4 given that ∅ actually is 1/4. And we are taking it 3/4 when x=1. So, prob. of x=1 for Bin(1,1/4) is ¹C1×(1/4)¹×(3/4)^0 that is 1/4.
Thanks for the reply to this post. It is helpful. If I can, I have a follow up question... The value of ∅ can be one of three values: 1/4, 1/2 or 3/4. So when using the Bin (1,∅) distribution why do we use p=∅ when there is only a 1 in 3 chance of ∅ occuring (or a 2/3 chance of it not occuring)? (Hope this question makes sense.) Thanks
