# 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. Commented Nov 23, 2012 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. Commented Nov 23, 2012 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. Commented Nov 23, 2012 at 0:59
• cardinal is right - this is wrong. Commented Nov 23, 2012 at 1:19
• Okay, now it's fine. Commented Nov 23, 2012 at 5:45
• There's no need to reproduce the wrong answer. You can just write the correct answer & anyone who wants to read earlier drafts can look at the edit history.
– Sycorax
Commented Apr 23, 2023 at 21:40