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I am studying for the P/1 actuarial exam, and I keep encountering one type of problem which I can't seem to solve and can't seem to find any generalized explanation for how to do it in my textbooks. The problems always involve multiple random variables with identical distributions, and the solution always involves the $max \{x_1, x_2, .. x_n\}$ function.

Here are three examples of this type of problem. Note that I am not looking for specific solutions to each problem. Such solutions are extant. I am trying to figure out if there is a general procedure or process I can use to attack this kind of problem.

Example 1:

A system made up of 7 components with independent, identically distributed lifetimes will operate until any of 1 of the system' s components fails. If the lifetime X of each component has density function: $$f(x)= \begin{cases}\lambda {(\frac {3}{x})}^4 & x > 1 \\ 0 & \text{otherwise} \end{cases}$$

What is the expected lifetime until failure of the system?

Example 2:

A game is played where a player generates 4 independent Uniform(0,100) random variables and wins the maximum of the 4 numbers. (a) Give the density function of a player's winnings. (b) What is the player's expected winnings?

Example 3:

A company agrees to accept the highest of four sealed bids on a property. The four bids are regarded as four independent random variables with common cumulative distribution function $$F(x)= \frac{1} {2} (1+sin \pi x),~ \frac {3}{2} \le x \le \frac{5}{2}$$

and 0 otherwise. What is the expected value of the accepted bid?

Personal note: I plan to take my 1/5 odds (a.k.a skip) on anything that involves trig identities, but I include this third example to round out the group. I am pretty sure they all use the technique in question.

Thank you for your enlightenment!

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I think the first example asks for min, not max. However, [this answer] (…) could be useful for expected value of max and you could adapt for min. . – Macavity Mar 14 '13 at 4:12
Yes, it is the min. Thank you for the link. In combo with the answer from @Andre below the process is very clear for me now. – Katarzyna Mar 14 '13 at 6:01
up vote 1 down vote accepted

The problems are not all about "max." The first one is about "min." The system fails as soon as a component fails. So its lifetime is the minimum of the lifetimes of the components.

Let $X_1,X_2, \dots,X_n$ be identically distributed independent random variables, with continuous distribution.

Let $Y$ be the minimum of the $X_i$, and let $Z$ be the maximum of the $X_i$. We want to find the distribution of $Y$ and of $Z$.

For any $i$, let $F$ be the cumulative distribution function of $X_i$ (they are all the same.)

The minimum: We have $Y\gt y$ if and only if all the $X_i$ are $\gt Y$. For any $i$, the probability that $X_i \gt y$ is $1-F(y)$. So the probability they are all $\gt y$ is $(1-F(y))^n$. Thus the probability that $Y\le y$ is equal to $1-(1-F(y))^n.$

If we can calculate $F$, we know the cdf of $Y$, and therefore, if we need it, the density function of $Y$. Once we have these, most common problems about $Y$ are reasonably straightforward.

The maximum: This is simpler. The maximum $Z$ is $\le z$ if and only if all the $X_i$ are $\le z$. the probability of this is $(F(z))^n$. So the cumulative distribution function of $Z$ is $(F(z))^n$.

All three of your problems can be solved using the facts about the distribution of $Y$ and $Z$ just derived.

One does not need to remember the formulas: in any particular case, we can repeat the reasoning that led to the formulas.

If you need the calculation for any specific one of the three problems, it can be added.

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Wow! Most excellent! This was EXACTLY what I needed. (As soon as my rep gets high enough I will come back and upvote this.) – Katarzyna Mar 14 '13 at 6:00
This gets better every time I read it. ;-) – Katarzyna Mar 23 '13 at 2:58

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