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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.

Many thanks in advance.

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    $\begingroup$ 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. $\endgroup$ – Elvis Jan 25 '12 at 13:03
  • $\begingroup$ 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 $\min \{X_i \>:\> Y_i = 1 \}$? $\endgroup$ – Elvis Jan 25 '12 at 13:11
  • $\begingroup$ @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. $\endgroup$ – cardinal Jan 25 '12 at 14:30
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I take the following interpretation of your question:

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 $X_{(i)} = X_j$).

\begin{align*} \mathbb P(M = k) =& \mathbb P( X_{(1)}=k,\ Y_{(1)} = 1) + \mathbb P( X_{(2)}=k,\ Y_{(1)} = 0, Y_{(2)} = 1 ) + \cdots \cr &\cdots + \mathbb P( X_{(n)}=k,\ Y_{(1)} = 0, Y_{(2)} = 0, \dots, Y_{(n)} = 1) \cr =& p \cdot \mathbb P(X_{(1)}=k) + (1-p)p\cdot \mathbb P(X_{(2)}=k) + \cdots\cr &\cdots + (1-p)^{n-1} p \mathbb P(X_{(n)}=k) \cr =& p\cdot\sum_i (1-p)^{i-1} \mathbb P(X_{(i)} = k) \cr \mathbb P(M = +\infty) =& (1-p)^n \end{align*}

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.

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