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This is from an online course textbook (Harvard Statistics 110: see #45, p. 28 of pdf). It's not a homework question, as the answer is provided in the text. I just really want to understand why the solution is that way.

A new treatment for a disease is being tested, to see whether it is better than the standard treatment. The existing treatment is effective on $50\%$ of patients. It is believed initially that there is a $2/3$ chance that the new treatment is effective on $60\%$ of patients, and a $1/3$ chance that the new treatment is effective on $50\%$ of patients. In a pilot study, the new treatment is given to $20$ random patients, and is effective for $15$ of them.

(a) Given this information, what is the probability that the new treatment is better than the standard treatment?

(b) A second study is done later, giving the new treatment to $20$ new random patients. Given the results of the first study, what is the PMF for how many of the new patients the new treatment is effective on? (Letting $p$ be the answer to (a), your answer can be left in terms of $p$.)


The answer to (b):
Let $Y$ be how many of the new patients the new treatment is effective for and $p=P(B|X=15)$ be the answer from (a). Then for $k\in \{0,1,\ldots,20\}$,
\begin{align} P(Y=k|X=15) &= P(Y=k|X=15,B)P(B|X=15) \\ &\quad + P(Y=k|X=15,B^c)P(B^c|X=15) \\[5pt] &= P(Y=k|B)P(B|X=15) + P(Y=k|B^c)P(B^c|X=15) \\[5pt] &= {20\choose k} (0.6)^k(0.4)^{20-k}p + {20\choose k}(0.5)^{20}(1-p). \end{align}

Questions:

  1. Before I looked at the answer, I didn't think that the answer to (a) would be found using a binomial distribution. But after I read the answer I was convinced by it. So, is there a kind of "general hint" when to use a binomial distribution?

  2. I don't understand why the answer to (b) is that way. Specifically, why is $$P(Y=k|X=15, B)=P(Y=k|B)$$ and the analogous one conditioning on $B^c$?
    This equality is implied from the second equality of (b)'s answer. $$P(Y=k|\underbrace{X=15, B}_{\text{intersection removed}})P(B|X=15)+P(Y=k|\underbrace{X=15, B^c}_{\text{the same here}})P(B^c|X=15)$$ $$=P(Y=k|B)\ \ \ \ \ \ P(B|X=15)\ \ \ \ \ + \ \ \ \ \ \ \ \ P(Y=k|B^c) \ \ \ \ \ P(B^c|X=15)$$

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2 Answers 2

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Specifically, why is $$P(Y=k|X=15, B)=P(Y=k|B)$$

  • The expression P(Y=k|X=15,B) = P(Y=k|B) is allowed because

    $$P(Y=k|X=i,B) = {{20}\choose {k}} 0.6^k0.4^{20-k} $$

    independent from $X=i$, so it can be left out

    This logic is indeed not generally true, ie $P(a|b,c)$ may be different from $P(a|c)$, and is just true for this specific case where the experiments are independent (new random patients) and it is assumed that the result have no correlation other than that they are both defined by the condition $B$ or $B^c$.

    Comparable expression with a dice roll for any $k$ and $l$: $$P(\text{next roll} =k \, | \,\text{ last roll} = l, \text{ dice = fair })=\frac{1}{6}$$

    when we assume the rolls are independent. For that dice roll you are allowed to write $P(k|l,fair) = P(k|fair) = 1/6$

Before I looked at the answer, I didn't think that the answer to (a) would be found using a binomial distribution. But after I read the answer I was convinced by it. So, is there a kind of "general hint" when to use a binomial distribution?

  • Effectively you have a mixture distribution of two binomial distributions

    $$ p \underbrace{ {{20} \choose {k}} 0.6^k0.4^{20-k}}_{\text{Binom$(20,0.6)$}} + (1-p) \underbrace{ {{20} \choose {k}} 0.5^k0.5^{20-k}}_{\text{Binom$(20,0.5)$}} $$

    being equivalent to a probability $p$ that the medicine works on 60% of the people and a probability $1-p$ that the medicine works on 50% of the people.

  • This logic may not seem so correct. How could one, based on question 'a', believe that you have the medicine working on only either 50% or 60%? That is a false dichotomy.

    However, note that this is created due to the weird prior believe that it is either working on 50% or 60% (probabilities 2/3 vs 1/3) and nothing else (so the posterior distribution will stick to this weird believe and only change these ratios 2/3 and 1/3 into $p$ and $1-p$).

    It is not a realistic question, what one would get in practice. But it illustrates a more complicated principle (see next point).

  • Eventually (a better alternative to the question problem statement which may occur later in your course), one could use a beta-binomial distribution in which every probability has an assigned value (not just 2/3 60% and 1/3 50%, or $p$ 60% and $1-p$ 50%). Then the posterior distribution for $f$, the fraction of people on which the medicine works (which could be modeled as a beta distribution), is a spectrum instead of just two values. This spectrum of probabilities for different $f$ is then used to write out the beta-binomial distribution.

  • note that the letter $p$ is often a parameter for the binomial distribution, the probability that the medicine works. Here it is used differently for the probabilities $p$ and $1-p$ that the medicine works with 60% or 50%. This might be confusing.

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  • $\begingroup$ Many thanks for your answer :); it's detailed and convincing. I hope the following illustrate what I meant by my expression. In the answer to (b)[the image]: $$P(Y=k|\underbrace{X=15, B}_{\text{intersection taken off}})P(B|X=15)+P(Y=k|\underbrace{X=15, B^c}_{\text{the same here}})P(B^c|X=15)$$ $$=P(Y=k|B)\ \ \ \ \ \ P(B|X=15)\ \ \ \ \ + \ \ \ \ \ \ \ \ P(Y=k|B^c) \ \ \ \ \ P(B^c|X=15)$$ $\endgroup$
    – Abdu Magdy
    Mar 28, 2018 at 10:15
  • $\begingroup$ @AbduMagdy the probability is $$P(Y=k|X=i,B) = {{20}\choose {k}} 0.6^k0.4^{20-k} $$ independent from i. So you can leave that part out (and similarly for the other expression with $B^c$). ----- Was this (the step in which X=15 gets dropped from the condition) the heart of the matter? (should I change my answer to focus on this point?) Or was it about the entire derivation? $\endgroup$ Apr 17, 2018 at 22:32
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In this response I will limit myself to the general question of when to use the binomial.

The binomial arises naturally as the count of successes in a sequence of independent Bernoulli trials with constant success probability $p$.

It's useful to keep the conditions in mind (both for Bernoulli trials and the independent and constant $p$ of the sequence of them) and see whether that might be more-or-less plausible. [Often the model is a reasonable approximation when the conditions don't quite hold.]

For example if you have a population with some fraction having some attribute and you sample it at random with replacement, the count of the number having the attribute in your sample would be an obvious candidate for using a binomial. If you instead sample without replacement but the population is very very large compared to your sample (making the effect on $p$ at trial $i$ and the dependence negligible) then you could use the binomial as an approximation.

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  • $\begingroup$ Thanks so much for you answer :). The hardship really arises (at least from my viewpoint) when you try to apply the characteristics you've provided on the experiment in hand. For instance, the treatment experiment in the OP is so convoluted and confusing that I couldn't specify (recognize or identify) the Bernoulli Trials and $p$ that I can build on my model. However, I think it needs more experience. $\endgroup$
    – Abdu Magdy
    Mar 29, 2018 at 20:27

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