0

I want to derive the distribution of $M_n=min(X_1 ... X_n)$ in another way than by a combinatorial analysis. Say we have $X_1...X_n$ represent $n$ draws without replacement from the numbers $1...N$ with equal equality.

First, I want to compute: $P(M_n \leq k)=P(X_1 \leq k) \cdot ... \cdot P(X_n \leq k)$. If $k>n$ we want to get $n$ numbers less or equal to number $k$. So the probability $P(M_n \leq k) = \frac{{k\choose n}}{N \choose n}$.

Second, I want to derive $P(M_n = n)$. Clearly this is equal to $\frac{1}{N \choose n}$.

Now I want to show that for $k=n+1, ..., N$ $P(M_n = k) = n\cdot \frac{(k-1)!}{(k-n)!} \cdot \frac{(N-n)!}{N!}$. And compute from this the distribution of $M_n$.

iJup
  • 1,999

1 Answers1

1

Notice that:

$\{M_n=k\}\equiv \{k \,\, \text{be} \,\, \text{in the sample and the others be greater than}\,\, k\}$

and

$\# \{\text{greater than}\,\, k\}=N-k$

therefore

$P(M_n=k)=\dfrac{\binom{1}{1}\binom{N-k}{n-1}}{\binom{N}{n}}=\dfrac{\binom{N-k}{n-1}}{\binom{N}{n}}$.

savalan savalan
  • 71