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Suppose $X_i$ is an indicator random variable. There is another random variable Z defined as $Z = \min(c, \sum_i X_i)$, where $c$ is a constant. How do we compute $E[Z]$? I have come up with the following expression, but I am not sure if its correct, $E[Z] = \min(c, \sum_i Pr(X_i = 1))$.

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There is no simple formula and the one you suggest is incorrect, as can be seen readily by checking some simple cases. –  Did Jun 19 '12 at 19:08
    
Can we atleast claim that $E[Z] \leq \min(c, \sum_i Pr(X_i = 1))$ ? –  Shalmoli Gupta Jun 20 '12 at 10:06
    
Yes: $Z=\min(c,S)$ implies $E(Z)\leqslant\min(c,E(S))$. –  Did Jun 20 '12 at 10:17

1 Answer 1

Assuming $X_i$ are iid with $p=P(X_i=1)$, then $Y=\sum_{i=1}^n X_i$ follows a $(n,p)$ Binomial distribution. Calling $A=A(n,p,c)=P(Y < c)$, $Z$ is the mixing of a truncated Binomial and a discrete Dirac delta (I'm assuming $c$ is an integer).

$$P(Z=z) = A {n \choose z} p^z (1-p)^{n-z} + (1-A) \delta(z-c) \hspace{1 cm} 0 \le z \le c$$

And so

$$E(Z) = A \sum_{z=0}^{c-1} z {n \choose z}p^z (1-p)^{n-z} + (1-A) \, c $$

with $$A=\sum_{z=0}^{c-1} {n \choose z}p^z (1-p)^{n-z}$$

I don't think this can be simplified much.

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I may not require an exact expression for $E[Z]$, Can we claim that $E[Z] \geq \min(c, \sum_i E[X_i]) = \min(c, \sum_i Pr(X_i = 1))$ ? –  Shalmoli Gupta Jun 20 '12 at 10:18
    
user: I do not get it: I answer that in the comments. –  Did Jun 21 '12 at 6:18

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