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Suppose there are $n$ dependent Bernoulli trials, $X_{1}$,...,$X_{n}$ with $% X_{j}\in \{1,0\}$ and $\Pr (X_{j}=1)=q$ for all $j=1,...,n$. For any $% n\geqslant 2$ dependent Bernoulli trials, in the most general case, defining a joint distribution of $\{X_{j}\}_{j=1}^{n}$ requires specification of $2^{n}-1$ parameters. As in Witt (2014), I make two assumptions to reduce the required parameter set to two: $(q,\rho )$, where, $q$ is already defined and $\rho $ is defined in Assumption 2.

ASSUMPTION 1: $(X_{1},\ldots ,X_{n})$ are exchangeable, that is, \begin{equation*} \Pr (X_{1}=\lambda _{1},\ldots ,X_{n}=\lambda _{n})=\Pr (X_{\pi (1)}=\lambda _{1},\ldots ,X_{\pi (n)}=\lambda _{n}), \end{equation*} where $\lambda _{1},\ldots ,\lambda _{n}\in \{0,1\}$ and $\pi $ is a permutation of $\{1,\ldots ,n\}.$

ASSUMPTION 2: Constant pairwise correlation, that is, \begin{equation*} \rho =\operatorname{Corr}(X_{i},X_{j}|X_{\pi (1)}=1,\ldots ,X_{\pi (k)}=1)>0 \end{equation*} for $k=0,\ldots ,n-2$ and for any permutation $\pi (1),\ldots ,\pi (k)$ of $k $ terms of the sequence of $n-2$ integers $1,\ldots ,n$ excluding $i$ and $j$

With these two assumptions, Witt (2014) shows \begin{equation} \Pr (X_{j}=1|X_{\pi (1)}=1,\ldots ,X_{\pi (j-1)}=1)=1-\left( 1-q\right) \left( 1-\rho \right) ^{j-1}. \end{equation}

QUESTION: To show that \begin{equation*} \frac{\partial }{\partial \rho }\frac{\Pr (X_{k+1}=0,...,X_{n}=0|X_{1}=1,\ldots ,X_{k}=1)}{\Pr (X_{k+1}=0,...,X_{n}=0|X_{1}=0,\ldots ,X_{k}=0)}<0. \end{equation*}

My strategy is to show that $\frac{\partial }{\partial \rho }\Pr (X_{k+1}=0,...,X_{n}=0|X_{1}=1,\ldots ,X_{k}=1)<0$ and $\frac{\partial }{% \partial \rho }\Pr (X_{k+1}=0,...,X_{n}=0|X_{1}=0,\ldots ,X_{k}=0)>0$. Intuitively it makes sense that when Bernoulli trials are positively correlated, as the correlation increases, the probability that the next $n-k$ trials will be failures given that the first $k$ trials are successes should decrease. Conversely, as correlation increases, the probability that the next $n-k$ trials will be failures given that the first $k$ trials are also failures should increase. However, how do I prove it mathematically?

I can prove that $\frac{\partial }{\partial \rho }\Pr (X_{k+1}=0|X_{1}=1,\ldots ,X_{k}=1)<0$, by showing that $\Pr (X_{k+1}=0|X_{1}=1,\ldots ,X_{k}=1)= \left( 1-q\right) \left( 1-\rho \right) ^{k}$, which decreases with $\rho $.

But I am getting stuck at proving $\frac{\partial }{\partial \rho }\Pr (X_{k+1}=0,X_{k+2}=0|X_{1}=1,\ldots ,X_{k}=1)<0.$

I will appreciate any help.

[Please note that I have checked all related threads, but could not find any useful help].

References:

Witt, G., 2014. A Simple Distribution for the Sum of Correlated, Exchangeable Binary Data. Communications in Statistics-Theory and Methods, 43(20), pp.4265-4280.

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  • $\begingroup$ +1 But did you notice that "$p$" and "$\rho$" are rendered almost identically? Ouch! $\endgroup$ – whuber Feb 13 at 13:40

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