# Probability of a random variable to be the largest among others

Let us have $N$ random variables generated by uniform distribution. That is, $$u_i \sim \mathcal{U}(0,1),\quad i=1,\ldots,N$$.

What is the probability of $u_N$ being the largest? I.e., how can I compute $$p\Big(u_N\geq \max(u_1,u_2,\ldots,u_{N-1})\Big)$$

• Please see my comment to Tenali. The correct answer is independent of the distribution as long as it is continuous and the random variables are iid. – cardinal Nov 23 '12 at 1:00

If the $u_i$ are independent, the probability is $\frac{1}{N}$ for any continuous distribution common to all the $u$'s.

This is obvious from basic symmetry considerations.

Is this homework?

• No, it's not a homework question. I wondered how it would be to get a max probability. – Helen Nov 23 '12 at 2:06

[wrong] $$P(max(u_1, \ldots, u_{N-1}) \leq u_N) = P(\cap_{i = 1}^{N-1} (u_i \leq u_N)) = \prod_{i = 1}^{N - 1} P(u_i \leq u_N) = P(u_1 \leq u_N)^{N - 1} = \frac{1}{2^{N - 1}}$$ [\wrong]

As has been identified in the comments, this is the wrong approach to take. Sorry, solving probability problems in the dead of the night is bad for reputation :P.

Let $A_i$ be the event that $max(u_1, \ldots, u_{i-1}, u_{i+1}, \ldots, u_N) < u_i$. The $A_i$'s are disjoint by design and by symmetry, all the $A_i$'s have the same probability. Therefore $P(A_N) = \frac{1}{N}$.

• Unfortunately, this is not correct. The events $\{u_i \leq u_N\}$ are not independent. Instead, use symmetry. – cardinal Nov 23 '12 at 0:59
• cardinal is right - this is wrong. – Glen_b Nov 23 '12 at 1:19
• Okay, now it's fine. – Glen_b Nov 23 '12 at 5:45