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Let A,B,C,D,E be independent random variables, each of which is uniformly distributed in the interval [0,18]. Let X=min{A,B,C,D,E}. What is the expected value of X? Enter your answer as a decimal.

I know that I need to be attempting some work,but any direction would be greatly appreciated!

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We find the density function of $X$. For $0\lt x\lt 18$, we have $X\gt x$ if all of $A,B,C,D,E$ are $\gt x$.

The probability that $A$ is $\gt x$ is $\frac{18-x}{18}$. So the probability that all of $A,B,C,D,E$ are $\gt x$ is $\left(\frac{18-x}{18}\right)^5$.

It follows that $$F_X(x)=1-\left(\frac{18-x}{18}\right)^5.$$ Now we can differentiate to find the density function $f_X(x)$, and then calculate $E(X)$ as usual. The substitution $u=18-x$ will be useful for the evaluation of the integral. Or else you can use integration by parts.

Remark: There is a "fancier" way that bypasses finding the density function, if you happen to know an expression for the expectation in terms of the cdf.

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  • $\begingroup$ Does it make sense that the integral would be x*(5/18)*((18-x)/18)^4dx evaluated from 0 to 18? $\endgroup$ – user130272 Apr 8 '14 at 5:16
  • $\begingroup$ Yes, that is a correct expression for the expectation. $\endgroup$ – André Nicolas Apr 8 '14 at 5:18
  • $\begingroup$ Thanks for all of the help! $\endgroup$ – user130272 Apr 8 '14 at 5:23
  • $\begingroup$ You are welcome. Numerically, the answer will be quite simple. $\endgroup$ – André Nicolas Apr 8 '14 at 5:24
  • $\begingroup$ Okay integrating I got 3! Thanks! $\endgroup$ – user130272 Apr 8 '14 at 5:31
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Let's suppose a set of identically uniformly distributed random variables $A,B,C,D,E$ and assume that we want to find the distribution $X=\min{\{A,B,C,D,E\}}$.
Then the probability $P(X≤x)=P(\min{\{A,B,C,D,E\}}≤x)$ implies that one can find at least one of the elements of $\{A,B,C,D,E\}$ smaller than $x$.
Or one can say that the probability that at least one of them is smaller than $x$ equals to: $$\begin{eqnarray}P(\min{\{A,B,C,D,E\}}≤x)&=&1-P(A>x,B>x,C>x,D>x,E>x)=\\&=&1-P(A>x)P(B>x)P(C>x)P(D>x)P(E>x) =\\&=&1-\left(P(Y>x)\right)^5 \end{eqnarray}$$ where $Y$ is any of the variables $\{A,B,C,D,E\}$ and $F_Y(y)$ its cdf (since they are identically distributed.)
Then one can conclude that the cdf of $X$ would look like:$F_X(x)=1-(1-F_Y(x))^5.$
Since you have $Y\sim \text{Uniform} (0,18)$, then the cdf $F_Y=\frac{x}{18}$. Then $F_X=1-(1-\frac{x}{18})^5$.

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If you have a finite number of i.i.d random variable, their maximum, their minimum, and any $j$-rank is a RV with a closed form pdf and cdf. As such, it should be easy for you to find the expectation. These are called the rank statistics of a sample. Casella and Berger's "Statistical Inference" has a nice treatment of the subject.

Sorry I'm not giving an explicit answer, I just figured you would benefit more from learning it in a general context than from seeing an equation.

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