problem : After bikes are produced by the factory, they appear in a test center, where they are tested for defects. Only 20% are usually tested as defective . Let S1= number of bikes tested until first positive result, S2= number of bikes tested until second positive result. compute $E[S2|S1=n]$ and compute $E[S2]$ via de identity $E[S2] = E[ E[S2|S1] ]$
Could anyone help to understand how to calculate?
On
Your approach could work but the formulas you inserted contain some errors (if you used Math Jax they could be more understandable). If you wanna calculate the Conditional Expectation you can do in the following way:
$$\mathbb{E}[S_2|S_1=n]=\sum_{x=n+1}^{\infty}x p q^{x-n-1}=\sum_{x=n+1}^{\infty}[(x-n)+n] p q^{(x-n)-1}=\sum_{(x-n)=1}^{\infty}[(x-n)+n] p q^{(x-n)-1}=$$
$$\sum_{y=1}^{\infty}[y+n ]p q^{y-1}=\sum_{y=1}^{\infty}yp q^{y-1}+n\sum_{y=1}^{\infty}p q^{y-1}=\mathbb{E}[S_1]+n\times1=5+n$$
...but they are redundant calculations....the best way to proceed is to observe that, once that the first defective item is detected, the remaining items are "good as new" so the expectation is always $\frac{1}{p}=5$ plus the $n$ previous trials already effected.
You can use the lack of memory property of the Geometric distributioni finding immediately
$\mathbb{E}[S_2|S_1=n]=n+5$
....and now the result of the marginal $\mathbb{E}[S_2]$ is trivial
The other way to calculate $\mathbb{E}[S_2]=10$ is via Negative Binomial law