1
$\begingroup$

An urn contains balls numbered 1 to N. Let X be the largest number drawn in n drawings when random sampling with replacement is used. (The event X k means that each of n numbers drawn is less than or equal to k.) Show that when N is large:

$${E[X] {\approx} \frac{n * N}{n+1}}$$

Here is my approach:

$$P(X=k) = P(X \leq k) - P(X \leq k-1) $$

since $$P(X \leq k) = \frac{k^n}{N^n}$$ and $$P(X \leq k-1) = \frac{(k-1)^n}{N^n}$$ then $$E(X = k) = \sum_{k=1}^N\frac{k*k^n}{N^n} - \sum_{k=1}^N\frac{k*(k-1)^n}{N^n}$$

By using telescoping sum and Riemann integral, I am getting the answer:

$${E[X] {\approx} \frac{N}{n+1}}$$

where I don't have (n) term on the numerator.

What is wrong in my approach? Thanks.

Disclaimer: This is part of my "Probability" HW.

$\endgroup$
1
  • 1
    $\begingroup$ You should add the self-study tag. $\endgroup$
    – GeoMatt22
    Oct 19, 2016 at 2:58

1 Answer 1

1
$\begingroup$

Hint

If we call the random draws $U_1, U_2, \ldots, U_n$ then for each $i$ the quantity $U_i / N$ will behave increasingly like a continuous uniform$(0, 1)$ random variable as $N \to \infty$. What is the distribution of the maximum of $n$ uniform$(0, 1)$ random variables?

$\endgroup$
1
  • 1
    $\begingroup$ Just as I starting typing ... you beat me to it! $\endgroup$
    – GeoMatt22
    Oct 19, 2016 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.