Expected value of min X for bernoulli success?

I take a SRS sample of size n from a population of x values ranging from 1 to N. Each selected unit also has a probability p of success or q = 1-p of failure (i.e. the probability of success/failure is independent of the value of x).

What is the expected value of the minimum x value for which success occurs? I believe that when n = N I can just use the geometric distribution, but I can't deal with the problem once only a proportion of the units are sampled.

• What is "SRS"? And what is x?! Could you consider rewording a little bit your question? I don’t get it at all. Please consider formalizing it with math notations. – Elvis Jan 25 '12 at 13:03
• Do you mean your observations are $(X_i, Y_i)$ for $i = 1, \dots, n$, with $X_i$ drawn uniformly (with or without replacement?) in $\{1, \dots, N \}$ and $y_i$ drawn independently from a Bernoulli $\mathcal B(p)$, and you’re asking for the distribution of $\min \{X_i \>:\> Y_i = 1 \}$? – Elvis Jan 25 '12 at 13:11
• @Elvis: I had this question on this site too at one point. Apparently this is a common initialism in certain applied areas that use statistics, though not common at all in the statistics literature itself. SRS stands for simple random sample. – cardinal Jan 25 '12 at 14:30

The observations are $(X_i, Y_i)$ for $i = 1, \dots, n$, with $X_i$ drawn without replacement in $\{1, \dots, N \}$ and $y_i$ drawn independently from a Bernoulli $\mathcal B(p)$. You’re asking for the distribution of $$M = \min \{X_i \>:\> Y_i = 1 \}.$$ Note that $M$ can be equal to $+\infty$ as all $Y_i$ can be equal to 0.
Let us denote $X_{(i)}$ the $i$-th order statistic of the sample $X_1,\dots,X_n$ and $Y_{(i)}$ the $Y_i$’s reordered as the $X_i$’s (ie $Y_{(i)} = Y_j$ if $X_{(i)} = X_j$).
On the other hand, $\mathbb P(X_{(i)} = k)$ can be found just by counting the favorable events. There are $N \choose n$ ways to draw $n$ elements among $N$. If the $i$-th element is $k$, the $i-1$ first have to be drawn among $k-1$ elements, and the $n-i$ remaining among $N-k$ elements. Hence $$\mathbb P(X_{(i)} = k) = { {k-1 \choose i-1} {N-k \choose n-i} \over {N\choose n}} \hspace{1cm} (\text{for } i \le k \le i + N -n),$$ and we are done.