# Expected value of a max

We have a roulette with the circumference $a$. We spin the roulette 10 times and we measure 10 distances, $x_1,\ldots,x_{10}$, from a predefined zero-point. We can assume that those distances are $U(0,a)$ distributed.

An estimation of the circumference $a$ is given:

$$a^* = \max(x_1,\ldots,x_{10})$$

To check whether it's biased or not I need to calculate:

$$E(a^*) = E(\max(x_1,\ldots,x_{10}))$$

How do I proceed? I don't know any rules for calculating the estimate of a $\max$.

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Order Statistics? – Inquest Dec 13 '12 at 16:32
@Inquest ??  – Did Dec 13 '12 at 16:37
@did, isn't what Andre Nicolas suggested called Order Statistics? – Inquest Dec 13 '12 at 16:38
@Inquest Not really. Usually, one fers to order statistics when the joint distribution of the ordered sample is involed, not just the maximum of the sample. – Did Dec 13 '12 at 18:46

Let $W=\max(X_1,\dots,X_{10})$. Then $W\le w$ if and only if $X_i\le w$ for all $i$. From this you can find the cdf of $W$, hence the density, hence the expectation.

Added: For any $i$, the probability that $X_i\le w$ is $\dfrac{w}{a}$.

So by independence, the cumulative distribution function $F_W(w)$ of $W$ is $\left(\dfrac{w}{a}\right)^{10}$ (for $0\le w\le a$)

It follows that the density function of $W$ is $\dfrac{1}{a^{10}}10w^9$ on $[0,a]$, and $0$ elsewhere.

Multiply this density function by $w$, integrate from $0$ to $a$ to find $E(W)$.

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The detour by the densities is not needed. – Did Dec 13 '12 at 16:36
Certainly true. It is a matter of guessing what is likely the more comfortable path for the student in a typical introductory course. – André Nicolas Dec 13 '12 at 16:39
Sorry but I don't understand your answer. What is $w$? Where did you get it from? Is "iff" a typo or some special notation? – Dimme Dec 13 '12 at 16:46
$w$ is a variable. Random variable $W$ is the maximum. I am using $n$ instead of $10$. Have expanded the "iff". The cdf of $W$, by hint given above, is $(w/a)^n=(1/a^n)w^n$ (on $[0,a]$). Density is therefore $(1/a^n)nw^{n-1}$. For expectation, multiply by $w$, integrate from $0$ to $a$. Get $\dfrac{na}{n+1}$, in your case $\dfrac{10a}{11}$. So estimator not unbiased, can make it unbiased by multiplying by $11/10$. – André Nicolas Dec 13 '12 at 17:17
I have done some more exercises now and I understand more. You answer was actually quite helpful. I have one more question if you don't mind: Why is E(W) the integral of f(w) * w? I don't seem to find a formula that looks like that in my book. – Dimme Dec 14 '12 at 0:50

Hint: (1) Find $\mathbb P(a^*\lt t)$ for every $t$ in $(0,a)$. (2) Find a formula for $\mathbb E(a^*)$ as a function of the probabilities $\mathbb P(a^*\lt t)$.

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I guess that $P(a^* < x) = 0$ for any $x$ since $a^*$ is estimated to be the biggest $x$ out there and no other $x$ can be bigger that the biggest. So far so good, but I don't understand your second hint. How can an expected value be a function of probabilities 0? – Dimme Dec 13 '12 at 16:42
You guessed incorrectly. Here $x$ is a running argument in $(0,a)$. Note that $a^*$ is not estimated and that the biggest $x$ means nothing. I modified the notations. – Did Dec 13 '12 at 18:48