相依离散均匀分布随机变量最小值的期望求解

Great question! Let's break down how to get that concise expression for ( E[Y] ), where ( Y = \min{X_1, ..., X_k} ) from drawing k balls without replacement from n numbered balls.

Step 1: Use the Alternative Expectation Formula for Non-Negative Integer Random Variables #

For any non-negative integer-valued random variable ( Z ), we can compute its expectation as:
[ E[Z] = \sum_{i=0}^{\infty} P(Z > i) ]
Since ( Y ) is a positive integer (it's the minimum of k numbered balls), this formula simplifies our work significantly—we don't need to compute ( P(Y = y) ) for each y and sum ( yP(Y=y) ).

Step 2: Simplify ( P(Y > i) ) #

As you noted, ( P(Y > i) ) is the probability that all k drawn balls have numbers greater than i. This is equivalent to choosing k balls from the ( n - i ) balls numbered ( i+1 ) to ( n ), divided by the total number of ways to choose k balls from n:
[ P(Y > i) = \frac{\binom{n-i}{k}}{\binom{n}{k}} ]
This is zero when ( n - i < k ) (i.e., ( i > n - k )), since we can't choose k balls from fewer than k options.

Step 3: Sum Using the Hockey-Stick Identity #

Now substitute this into the expectation formula:
[ E[Y] = \sum_{i=0}^{n-k} \frac{\binom{n-i}{k}}{\binom{n}{k}} ]
Let's make a substitution: let ( m = n - i ). When ( i = 0 ), ( m = n ); when ( i = n - k ), ( m = k ). Reversing the order of summation (which doesn't change the total), we get:
[ E[Y] = \frac{1}{\binom{n}{k}} \sum_{m=k}^{n} \binom{m}{k} ]
Here's where the hockey-stick identity comes in: this identity tells us that the sum of binomial coefficients ( \binom{k}{k} + \binom{k+1}{k} + ... + \binom{n}{k} ) equals ( \binom{n+1}{k+1} ). In formula form:
[ \sum_{m=k}^{n} \binom{m}{k} = \binom{n+1}{k+1} ]

Step 4: Simplify the Ratio of Binomial Coefficients #

Substitute the identity back into the expectation:
[ E[Y] = \frac{\binom{n+1}{k+1}}{\binom{n}{k}} ]
Now compute this ratio. Recall that ( \binom{a}{b} = \frac{a!}{b!(a-b)!} ):
[ \frac{\binom{n+1}{k+1}}{\binom{n}{k}} = \frac{\frac{(n+1)!}{(k+1)!(n+1 - (k+1))!}}{\frac{n!}{k!(n - k)!}} = \frac{(n+1)! \cdot k! \cdot (n - k)!}{n! \cdot (k+1)! \cdot (n - k)!} ]
Cancel out common terms (( n! ), ( k! ), ( (n - k)! )):
[ \frac{n+1}{k+1} ]

Final Result #

The concise expression for the expected value of the minimum drawn number is:
[ \boxed{E[Y] = \frac{n+1}{k+1}} ]

Let's verify with a quick example to be sure:

• If ( k = n ): We draw all balls, so ( Y = 1 ). The formula gives ( \frac{n+1}{n+1} = 1 ), which is correct.
• If ( k = 1 ): We draw one ball uniformly at random, so the expectation is ( \frac{n+1}{2} ), which matches the formula.
• If ( n = 4 ), ( k = 2 ): The expected value is ( \frac{5}{3} ), which matches the manual calculation of all possible draws.

内容的提问来源于stack exchange，提问作者M. BePe