Thanks man. You're awesome!
Let's see if I got it.
I will use the \Theta greek letter instead of a k because that was the one used on the original question.
1) Let (X_1,X_2,...X_n) be a random sample. The individual X_i are independent identically distributed random variables that follow an Uniform distribution X_i ~ U(0,\Theta)
2) The cumulative distribution function of the maximum is, by definition:
F(t) \equiv P(M \le t)
3) If the maximum value is \le t, that means all of the variables are, so:
P(M \le t) = P(X_1 \le t \; \cap \; X_2 \le t \; \cap \; \dots X_n \le t)=\prod_{i=1}^n\!P(X_i \le t)
The product follows because the individual X_i are independent random variables.
4) The probability density of the above Uniform distribution is:
f(x)=\frac{1}{\Theta} \; , \; 0 \le x \le \Theta
5) So it's distribution function is:
F(t)=0 \; , \; t \le 0
F(t)=\frac{t}{\Theta} \; , \; 0 < t \le \Theta
F(t)=1 \; , \; t > \Theta
6) So the product:
\prod_{i=1}^n\!P(X_i \le t)
equals the product of n distribution functions defined in point 5) which yields:
F(t)=0 \; , \; t \le 0
F(t)=(\frac{t}{\Theta})^n \; , \; 0 < t \le \Theta
F(t)=1 \; , \; t > \Theta
And the distribution of the maximum was found!
7) By differentiating the distribution function of the maximum one gets it's density function:
f(t)=\frac{nt^{n-1}}{\Theta^n} \; , \; 0 \le t \le \Theta
8) Finally the expected value of the max is:
E[max(X_1,X_2,...,X_n)]=E[t]=\int_{-\infty}^{+\infty}tf(t) \, dt=\int_0^{\Theta}t\frac{nt^{n-1}}{\Theta^n} \, dt=\frac{n}{\Theta^n}\int_0^{\Theta}t^n \, dt=\frac{n}{n+1}\Theta
So the result is neither \Theta nor \Theta \over 2 but a value that depends on the sample size and lies between these two. It will be exactly \Theta \over 2 when the sample size equals unity and it tends to \Theta when the sample size approaches infinity. This result is according to my original intuition so I think it is correct.
What do you think?
