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I have a random variable $1_{\{az_1+bz_i<L\}}$, 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 $X = \frac{\sum_{i=1}^n1_{\{az_1+bz_i<L\}}}{n}$, 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. – mike Jun 15 '12 at 12:06
@mike : But the indicator functions being summed are not independent, since each of them involves $z_1$. – Michael Hardy Jun 15 '12 at 13:36
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. – Michael Hardy Jun 15 '12 at 13:52
@mike, you are right that's exactly the problem, I need the detailed solution. – Vaolter Jun 15 '12 at 15:07
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 $\int \Phi(\frac {L - az}b)^{n-1} \phi(z) dz$. – mike Jun 15 '12 at 17:47

Introducing $L(z)=L-az$ and, for every $x$, $x(z)=x-\mathbf 1_{\{(a+b)z\leqslant L\}}$, the independence of $Z_1$ and $(Z_i)_{i\geqslant 2}$ yields $$P(X=x\mid Z_1=z)=P\left(\sum\limits_{i=2}^n\mathbf 1_{\{bZ_i\leqslant L(z)\}}=x(z)\right). $$ Since $(Z_i)_{i\geqslant 2}$ is i.i.d., this is the weight at $x(z)$ of the binomial distribution with parameters $n-1$ and $p(z)$, where $$ p(z)=P(bZ_i\leqslant L(z))=\Phi(L(z)/b). $$ To sum up, if $(a+b)z\gt L$, then for every integer $0\leqslant x\leqslant n-1$, $$P(X=x\mid Z_1=z)={n-1\choose x}p(z)^x(1-p(z))^{n-1-x}, $$ while, if $(a+b)z\leqslant L$, then for every integer $1\leqslant x\leqslant n$, $$P(X=x\mid Z_1=z)={n-1\choose x-1}p(z)^{x-1}(1-p(z))^{n-x}, $$

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