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I am attempting to answer the following question from my actuarial exams:

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We are asked to find $P(p=0.5|X=7)$.

Using Bayes Rule, I believe this is

$$P(p=0.5|X=7) = \frac{P(X=7|p=0.5)P(p=0.5)}{P(X=7|p=0.5)P(p=0.5) + P(X=7|p \sim Unif[0.5,1])P(p \sim Unif[0.5,1])}$$

Straightforwardly, using $X \sim Bin(8,p)$:

$$P(X=7|p=0.5)P(p=0.5) = {8 \choose 7}0.5^70.5 \times 0.8$$

I am less sure of $P(X=7|p \sim Unif[0.5,1])P(p \sim Unif[0.5,1])$, if even this is the right expression. Given it's continuous, I thought it might be:

$$P(X=7|p \sim Unif[0.5,1])P(p \sim Unif[0.5,1]) = \int^1_{0.5}{P(x|p)f(p) dp}$$

where $P(x|p) = {8 \choose 7}p^7(1-p)$ and $f(p) = 2$ (being the pdf of $Unif[0.5,1]$). But this doesn't account for the probability of p being uniformly distributed, $P(p \sim Unif[0.5,1]) = 0.2$.

What is the correct way to think about this?

Many thanks

Steven

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The question states the prior distributions on the hypotheses as $\pi(H_0)=0.8$ and $\pi(H_1)=0.2$. From Bayes' theorem: $$ p(H_0\,|\,k)=\frac{P(k\,|\,H_0)\pi(H_0)}{P(k\,|\,H_0)\pi(H_0) + P(k\,|\,H_1)\pi(H_1)} $$ We observe $k=7$ correct predictions out of $n=8$ total predictions. Using the binomial mass function, we can calculate the posterior as follows: $$ P(k\,|\,H_0) = {n\choose k}(0.5)^k(1 - 0.5)^{n-k} = {8\choose 7}(0.5)^7(1 - 0.5)^{1} = \frac{1}{32} $$ and for the alternative hypothesis, we integrate over its range, weighted by the density of the $U(1/2,1)$ distribution (which is $2$): $$ P(k\,|\,H_1) = \int_{0.5}^{1}2{n\choose k}(p)^k(1 - p)^{n-k}\,\mathrm{d}p = 2\int_{0.5}^{1}{8\choose 7}(p)^7(1 - p)^{1}\,\mathrm{d}p=\frac{251}{1152} $$ So finally we have: $$ p(H_0\,|\,7)=\frac{\dfrac{1}{32}0.8}{\dfrac{1}{32}0.8 + \dfrac{251}{1152}0.2} = \frac{144}{395} = 0.364557 $$

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  • $\begingroup$ I think you have left out the density of the uniform distribution of p from your integral... so p(k|H_1) is half of what it should be $\endgroup$ Aug 30 at 10:51
  • $\begingroup$ @StevenRJClarke1985 I think you're right, I have updated the calculations. Thanks for the heads up. $\endgroup$ Aug 30 at 14:19
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Great question! I think your denominator would be better written as

$$P(X=7|p=0.5)\cdot P(p=0.5) $$ $$+\int_{0.5}^1 P(X=7|p= r\cap 0.5<p<1 )\cdot P(p=r|0.5<p<1)\cdot P(0.5<p<1)dr$$.

Let me know if I have made a mistake. The term $P(p=r|0.5<p<1)$ is the pdf of the uniform distribution over (0.5,1) and the term $P(0.5<p<1)$ would account for the belief mass of 20%. Here is a related thread asking about how to apply these probability statements to the coin and octopus under investigation.

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