Probability

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.

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?

There seems to be something wrong with the typesetting.

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

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

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.

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??

1 Like

“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.

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

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 ?

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)

Got the same results as well

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.

Ex3.

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

Ex7.

  • Note that
    image
    and
    image
  • The conditional probabilities are therefore
    image
  • For one test being positive, the positive rate is

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

Q8-a)

Q8-b)

Q8-c)

The weighting should be highest for stocks with the most highly expected return.

Q8-d)

With no risk-free assets in a portfolio:

Please let me know if I am wrong.