Q&A 1.6
Here's the question: A statistician has to decide on the basis of a single observation from a Bin(1, theta) distribution whether the value of theta is ¼, ½ or ¾ . The loss suffered in the event of a wrong decision is 200 times the absolute value of the error made.
Solution
Loss matrix
Actual value of theta
1/4 1/2 3/4
0 50 100 (if statistician decides theta = 1/4)
50 0 50 (if statistician decides theta = 1/2)
100 50 0 (if statistician decides theta = 3/4)
For example, take the loss of 50 in the first row. This comes from 200 times the error made = 200 * (1/2 - 1/4) = 50.
Observations
The statistician must make a decision based on a single observation from
Bin (1, theta). Call that observation X. The values that X can take from this distribution are 0 and 1.
Now, according to a Bin(1, theta) distribution:
P(X = 0) = theta^0 x (1-theta)^1 = 1 - theta
and P(X = 1) = theta^1 x (1-theta)^0 = theta
So note, the lower theta is, the bigger the P(X = 0).
Decision table
Suppose the observation is X = 0. A sensible decision would be that theta = 1/4. Suppose the observation is X = 1. A sentible decision would be that theta = 3/4.
But this is just one of many decisions. In fact there are nine possible decisions:
Observation
X = 0 X = 1
1/4 1/4 (d1)
1/2 1/2 (d2)
3/4 3/4 (d3)
1/4 1/2 (d4)
1/4 3/4 (d5)
1/2 1/4 (d6)
1/2 3/4 (d7)
3/4 1/4 (d8)
3/4 1/2 (d9)
The sensible decision we referred to earlier is decision 5 (d5).
Risk function
A risk function is defined as the expected loss given a particular decision and a particlar actual value of theta. Call the risk function R.
R(d1, 1/4) = expected loss given the statistician makes decision 1 and the actual value of theta is 1/4.
How do we work out an expected loss? Expected loss is the sum of the losses times the associated probabilities.
If the statistician makes decision d1 then he decides (from the decision table) that theta is 1/4 if X = 0 and theta is 1/4 if X = 1.
If the actual value of theta is 1/4 then the loss the statistician makes (from the loss table) in both cases is 0.
So R(d1, 1/4) = 0 x P(X = 0) + 0 x P(X = 1)
= 0 x (1 - theta) + 0 x theta
= 0 x 3/4 + 0 x 1/4
= 0
Let's try another:
R(d4, 3/4) means the statistician has made decision 4 and the actual state of theta is 3/4.
If the statistician makes decision d4 then he decides (from the decision table) that theta is 1/4 if X = 0 and theta is 1/2 if X = 1.
If the actual value of theta is 3/4 then the loss the statistician makes (from the loss table) is 100 if X = 0 and 50 if X = 1.
So R(d4, 3/4) = 100 x P(X = 0) + 50 x P(X = 1)
= 100 x (1 - theta) + 50 x theta
= 100 x 1/4 + 50 x 3/4
= 62.5
There are 27 risk functions in total, since there are 9 decisions and 3 possible actual values of theta.
In the table below, I've highlighted in red the two numbers we've calculated.
The completed risk function table is as follows:
Actual value of theta
1/4 1/2 3/4
0 50 100 (d1)
50 0 50 (d2)
100 50 0 (d3)
12.5 25
62.5 (d4)
25 50 25 (d5)
62.5 25 12.5 (d6)
37.5 25 87.5 (d7)
75 50 75 (d8)
87.5 25 37.5 (d9)
Minimax
The max expected loss for each decision (reading from the above risk function table) is:
d1 100
d2 50
d3 100
d4 62.5
d5 50
d6 62.5
d7 87.5
d8 75
d9 87.5
The decisions that minimises the max expected loss are d2 and d5.
Bayes criterion solution
Prob (theta = 1/4) = 1/3
Prob (theta = 1/2) = 1/3
Prob (theta = 3/4) = 1/3
Actual value of theta
1/4 1/2 3/4
0 50 100 (d1)
50 0 50 (d2)
100 50 0 (d3)
12.5 25
62.5 (d4)
25 50 25 (d5)
62.5 25 12.5 (d6)
37.5 25 87.5 (d7)
75 50 75 (d8)
87.5 25 37.5 (d9)
To find the Bayes criterion solution, we need to work out the expected value of the risk function for each decision.
Eg, decision 1
E[R] = 0 x 1/3 + 50 x 1/3 + 100 x 1/3 = 50
etc
Then find the decision that minimises E[R]. This turns out to be a tie between decisions 2, 4, 5 and 6.
Are you sorted on this question now? Try the similar ones in the notes and in the X assignment and let me know.
Anna
