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I have a random variable $1_{\{az_1+bz_i, where $1_{\{.\}}$ is the indicator function, $z_1,z_i$ are $N(0,1)$, i.e., i.i.d standard normal and $L$ is a constant. If $X = \frac{\sum_{i=1}^n1_{\{az_1+bz_i, How do I determine the distribution function of $\widetilde{X}:=X|z_1 = z$, i.e., $X$ conditional on $z_1$. In other words, evaluate P$(\widetilde{X} \leq x)$.

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    conditional on $Z_1, \sum 1_{}$ is binomial, so you have a mixture of binomial, I haven't written down to see if you can explicitly do the mixing.2012-06-15
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    @mike : But the indicator functions being summed are not independent, since each of them involves $z_1$.2012-06-15
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    You've got a sum of Bernoulli-distributed random variables that are correlated. The correlations between pairs indexed by $i\neq j$ are all the same as each other if $i\neq 1\neq j$. The correlations between pairs indexed by $1$ and $i$ are all the same as each other. One hopes examining _pairwise_ dependences is enough, but I'm not sure it is.2012-06-15
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    @mike, you are right that's exactly the problem, I need the detailed solution.2012-06-15
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    I'm not sure where that discussion left us. Look at the sum, also, leave out the first, which has 2 $z_1$'s in it. The remainder is bin(n-1,p) with $p = \Phi(\frac {L - az_1}b)$. The probability of the sum being n-1 , for example is $\int \Phi(\frac {L - az}b)^{n-1} \phi(z) dz$.2012-06-15

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