Probability

MrBean · June 29, 2022, 8:40am

For the last question
If we assume the test result is deterministic, then

P(D2=1|D1=1) = 1
P(D2=0|D1=0) = 1

Doing first experiment twice does not add additional information. Therefore, P(H=1|D1=1,D2=1) == P(H=1|D1=1). You can derive the equation by doing some arithmetic.

timengler · August 20, 2022, 8:16am

I don’t understand equation 2.6.3 . On the right side, why wouldn’t P(A) on the top cancel out with P(A) on the bottom, and since the other term on the bottom right which is the sum of all b in B for P(B|A) equals 1, wouldn’t that mean it would then just simplify to P(A|B) = P(B|A) which is obviously incorrect?

Cesaryuan · August 24, 2022, 7:05am

There seems to be something wrong with the typesetting.

Gitartha_Kumar_Sarma · August 30, 2022, 6:46pm

Yes, the equation is slightly wrong. The updated equation should sum over all possible ‘a’ values in the sample space(a and its complement so that it gets normalized accurately), Reference

Denis_Kazakov · December 25, 2022, 9:11pm

An error in exercise 4? It says we draw n samples and then uses m in the definition of zm.

cajmorgan · January 23, 2023, 6:34am

Would have been great and more clear to actually see what numbers you multiplied in the example, for someone never doing stats before, it can be hard to comprehend all the formulas without any explanations.

tuntunia · February 18, 2023, 2:21pm

Answer for Problem 7:
Part 1:

                        | P(D2=1|H=0) | P(D2=0|H=0) | Total
P(D1=1|H=0)	|               0.02	|               0.08 | 0.10
P(D1=0|H=0)	|               0.08	|               0.82	| 0.90
Total                |               0.10 |                0.90 | 1.00

Part 2:

P(H=1|D1=1) = P(D1=1|H=1) * P(D2=0|H=1) / P(D1=1)

P(D1=1) = P(D1=1,H=0) + P(D1=1, H=1)
=> P(D1=1) = P(D1=1|H=0) * P(H=0) + P(D1=1|H=1) * P(H=1)

Thus, P(H=1|D1=1) = (0.99 * 0.0015) / ((0.10 * 0.9985) + (0.99 * 0.0015)) = 0.01465

Part 3:

P(H=1|D1=1,D2=1) = P(D1=1,D2=1|H=1) * P(H=1) / P(D1=1,D2=1)                     (1)
P(D1=1,D2=1|H=1) = P(D1=1|H=1) * P(D2=1|H=1)                                               (2)
P(D1=1,D2=1) = P(D1=1,D2=1,H=0) + P(D1=1,D2=1,H=1)
                        = P(D1=1,D2=1|H=0) * P(H=0) + P(D1=1,D2=1|H=1) * P(H=1)     (3)
Using (1), (2), & (3),
P(H=1|D1=1,D2=1) = 0.99 * 0.99 * 0.0015 / (0.02 * 09985 + 0.99 * 0.99 * 0.0015) = 0.06857

Are the above answers correct??

block_ramen · February 19, 2023, 6:20am

“For an empirical review of this fact for large scale language models see Revels et al. (2016).”

I believe this citation is wrong. It links to an auto-grad paper with nothing to do with evaluating LLMs.

Denis_Kazakov · March 25, 2023, 9:35am

I got the same answers independently. Now, we can try and calculate the probability of both of us being wrong.

SighingSnow · June 7, 2023, 8:51am

I have a qustion that, for the first problem of Q7 we have P(D1=1|H=0) = 0.1 but the condition listed above is P(D1=1|H=0) = 0.01 ?

Shawn_Shan · June 26, 2023, 5:55pm

You need to read carefully, it states: P(D1=0|H=1) = 0.01 (false negative). P(D1=1|H=0) = 0.1 (False positive)

Shawn_Shan · June 26, 2023, 6:01pm

Got the same results as well

cclj · July 20, 2023, 10:50am

The same answer. Quite conterintuitive, though. I think it is because the joint FPR 0.02 increased a lot compared to the original example 0.0003. So the positive result can still be confusing to patients.

cclj · July 20, 2023, 10:59am

Ex3.

The estimated probability is a random variable dependent on that follows the multinomial distribution:
- Expectation
- Variance
  
  scales . The convergence rate of is thus , consistent with the CLT.
According to the Chebyshev Inequality, one has

for a given deviation measurement .

Ex7.

Note that

and
The conditional probabilities are therefore
For one test being positive, the positive rate is

image692×136 10.4 KB

which is far from satisfactory since the false positive rate is too high.
For both two tests being positive,

image895×134 12.5 KB