posterior probability distribution

[logistic_guy]

Assume that the prior distribution for the proportion of defectives produced by a machine is

\(\displaystyle p\)​	\(\displaystyle 0.1\)​	\(\displaystyle 0.2\)​
\(\displaystyle \pi(p)\)​	\(\displaystyle 0.6\)​	\(\displaystyle 0.4\)​

Denote by \(\displaystyle x\) the number of defectives among a random sample of size \(\displaystyle 2\).

\(\displaystyle \bold{(a)}\) Find the posterior probability distribution of \(\displaystyle p\), given that \(\displaystyle x\) is observed.
\(\displaystyle \bold{(b)}\) Estimate the proportion of defectives being produced by the machine if the random sample of size \(\displaystyle 2\) yields \(\displaystyle 2\) defectives.

 

[khansaheb]

Assume that the prior distribution for the proportion of defectives produced by a machine is

\(\displaystyle p\)​	\(\displaystyle 0.1\)​	\(\displaystyle 0.2\)​
\(\displaystyle \pi(p)\)​	\(\displaystyle 0.6\)​	\(\displaystyle 0.4\)​

Denote by \(\displaystyle x\) the number of defectives among a random sample of size \(\displaystyle 2\).

\(\displaystyle \bold{(a)}\) Find the posterior probability distribution of \(\displaystyle p\), given that \(\displaystyle x\) is observed.
\(\displaystyle \bold{(b)}\) Estimate the proportion of defectives being produced by the machine if the random sample of size \(\displaystyle 2\) yields \(\displaystyle 2\) defectives.

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[logistic_guy]

Since the sample size \(\displaystyle n = 2\), and we are counting the probability of the number of defectives independently, then we have a binomial distribution. That is:

\(\displaystyle f(x|p) = \binom{2}{x}p^x(1 - p)^{2 - x}\)

where \(\displaystyle x\) is the number of defectives.

 

[logistic_guy]

The marginal distribution is:

\(\displaystyle g(x) = \sum f(x|p)\pi(p) = f(x|0.1)\pi(0.1) + f(x|0.2)\pi(0.2)\)


\(\displaystyle = \binom{2}{x}0.1^x(1 - 0.1)^{2 - x}0.6 + \binom{2}{x}0.2^x(1 - 0.2)^{2 - x}0.4\)


\(\displaystyle = \binom{2}{x}\bigg[0.1^x(0.9)^{2 - x}0.6 + 0.2^x(0.8)^{2 - x}0.4 \bigg]\)

 

[logistic_guy]

We know that \(\displaystyle x = 0,1,2\), then

\(\displaystyle g(0) = \binom{2}{0}\bigg[0.1^0(0.9)^{2 - 0}0.6 + 0.2^0(0.8)^{2 - 0}0.4 \bigg] = 0.742\)


\(\displaystyle g(1) = \binom{2}{1}\bigg[0.1^1(0.9)^{2 - 1}0.6 + 0.2^1(0.8)^{2 - 1}0.4 \bigg] = 0.236\)


\(\displaystyle g(2) = \binom{2}{2}\bigg[0.1^2(0.9)^{2 - 2}0.6 + 0.2^2(0.8)^{2 - 2}0.4 \bigg] = 0.022\)

And we have this table:

\(\displaystyle x\)​	0​	1​	2​
\(\displaystyle g(x)\)​	0.742​	0.236​	0.022​

 

[logistic_guy]

\(\displaystyle \bold{(a)}\) Find the posterior probability distribution of \(\displaystyle p\), given that \(\displaystyle x\) is observed.

The posterior probability distribution is given by:

\(\displaystyle \pi(p|x) = \frac{f(x|p)\pi(p)}{g(x)}\)

Then,

\(\displaystyle \pi(0.1|0) = \frac{0.1^0(0.9)^{2 - 0}0.6}{0.742} = \textcolor{blue}{0.655}\)


\(\displaystyle \pi(0.2|0) = \frac{0.2^0(0.8)^{2 - 0}0.4}{0.742} = \textcolor{red}{0.345}\)


\(\displaystyle \pi(0.1|1) = \frac{(2)0.1^1(0.9)^{2 - 1}0.6}{0.236} = \textcolor{blue}{0.458}\)


\(\displaystyle \pi(0.2|1) = \frac{(2)0.2^1(0.8)^{2 - 1}0.4}{0.236} = \textcolor{red}{0.542}\)


\(\displaystyle \pi(0.1|2) = \frac{0.1^2(0.9)^{2 - 2}0.6}{0.022} = \textcolor{blue}{0.273}\)


\(\displaystyle \pi(0.2|2) = \frac{0.2^2(0.8)^{2 - 2}0.4}{0.022} = \textcolor{red}{0.727}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\) Estimate the proportion of defectives being produced by the machine if the random sample of size \(\displaystyle 2\) yields \(\displaystyle 2\) defectives.

Let \(\displaystyle P\) be the proportion.

\(\displaystyle P = p_1\pi(0.1|2) + p_2\pi(0.2|2) = 0.1(0.273) + 0.2(0.727) = \textcolor{blue}{0.1727}\)