Using indicator variables, calculate the expected value

expected valueprobability

Its been a couple years since I took statistics, and I'm having a lot of trouble answering this question. I've been googling for the past three hours, and I don't feel any closer to figuring it out.

Consider a computer lab with 50 computers, at year 0. During any
subsequent year i, each computer may need to be individually replaced
with probability $\frac{i}{i+4}$. In each year, it is also possible that the entire
lab is replaced (due to disaster or sudden windfall of funds), with
probability $\frac{1}{10}$. Using indicator variables, calculate the expected > value (to 2 decimal places) for the number of the original machines that will > still be in the lab after 5 years.

Would I need indicator variables for both the case a machine needs to be replaced, and the case the whole lab is replaced?

Like…

     Ix{ 1, a machine needs replacement.
         0, otherwise. }

     Iy{ 1, entire lab needs replacement.
         0, otherwise. }

And how would I get the expected number of original machines after five years from this?

Best Answer

Let $X_i$ denote the indicator variable for the $i^{th}$ machine. That is, $X_i=1$ if the machine survives the five years and $X_i=0$ otherwise.

Then, clearly, $E[X_i]=E[X_j]$ for all $(i,j)$ so by Linearity the answer you want is $$E=E \left[ \sum X_i \right]=\sum E[X_i]=50\times E[X_1]$$

So we just need to compute $E[X_1]$, which is clearly $p_1$, the probability that this machine survives over the five years. In order to survive, there must be no decision to replace the entire lab over the five years...probability $\left(\frac 9{10}\right)^5$. Also, it must survive it's own individual failure process. That has probability $\left( 1- \frac i{i+4}\right)$ in year $i$. Thus $$p_1=\left(\frac 9{10}\right)^5\times \prod_{i=1}^5 \left( 1- \frac i{i+4}\right)=\frac {4374}{109375}\approx 0.0399909$$

Finally, the answer you want is $$E=50\times p_1=\frac {8748}{4375}\approx 1.99954$$

Thus, to the desired accuracy, you expect only $2$ machines to survive.

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