- $X_{1}, ... , X_{K}$ are $K$ independent Bernoulli random variables
- $c_{jk}$ are constants between 0 and 1.
- $p_{k} = P(X_{k} = 1)$ are known.
Is it possible to simplify the following expectation?
$$\sum_{\vec{x}} f(\vec{x}) P(\vec{X} = \vec{x}) $$
where
$$f(\vec{x}) = \prod_{j=1}^{J} \sum_{k=1}^{K} c_{jk} x_{k}$$
Thank you.
I can't generalize the case for $J > 2$.
Need a method to very quickly calculate / approximate the expectation for a J that is about 3000 and K is about 10.
Let :
- $A(i;j) = \sum_{k=1}^{K}c_{jk}p_{k}^{i}$
- $A(i;j_{1},j_{2}) = \sum_{k=1}^{K}c_{j_{1}k}c_{j_{2}k}p_{k}^{i}$
- $A(i;j_{1},j_{2},j_{3}) = \sum_{k=1}^{K}c_{j_{1}k}c_{j_{2}k}c_{j_{3}k}p_{k}^{i}$
For $J = 2$ and $K = 3$,
$\begin{split} f(\vec{x}) & = & \left (\sum_{k=1}^{3} c_{1k} x_{k} \right ) \left (\sum_{k=1}^{3} c_{2k} x_{k} \right) \\ & = & \sum_{k=1}^{3}\sum_{l=1}^{3} c_{1k} x_{k} c_{2l} x_{l} \end{split} $
$\begin{split} \sum_{\vec{x}} f(\vec{x}) P(\vec{X} = \vec{x}) & = &\sum_{x_{1} = 0}^{1}\sum_{x_{2} = 0}^{1} \sum_{x_{3} = 0}^{1} f(x_{1}, x_{2}, x_{3}) P(X_{1} = x_{1}, X_{2} = x_{2}, X_{3} = x_{3}) \\ &= &\sum_{x_{1} = 0}^{1}\sum_{x_{2} = 0}^{1} \sum_{x_{3} = 0}^{1} \sum_{k=1}^{3}\sum_{l=1}^{3} c_{1k} x_{k} c_{2l} x_{l} P(X_{1} = x_{1}, X_{2} = x_{2}, X_{3} = x_{3}) \\ &= & \sum_{k=1}^{3}\sum_{l=1}^{3} I(k \ne l) c_{1k} \cdot 1 \cdot c_{2l} \cdot 1 \cdot P(X_{k} = 1, X_{l} = 1) \\ &{}& + I(k = l) c_{1k} \cdot 1 \cdot c_{2k} \cdot 1 \cdot P(X_{k} = 1) \\ &= & \sum_{k=1}^{3}\sum_{l=1}^{3} I(k \ne l) c_{1k} c_{2l} p_{k} p_{l} + I(k=l) c_{1k} c_{2k} p_{k}\\ &= & \left [ \left(\sum_{k=1}^{3}c_{1k} p_{k}\right)\left(\sum_{l=1}^{3}c_{2l}p_{l} \right) - \sum_{k=1}^{3}c_{1k}c_{2k} p_{k} p_{k} \right ] + \sum_{k=1}^{3}c_{1k}c_{2k} p_{k}\\ &= & \left(\sum_{k=1}^{3}c_{1k} p_{k}\right)\left(\sum_{l=1}^{3}c_{2l}p_{l} \right) + \sum_{k=1}^{3}c_{1k}c_{2k} p_{k}(1 - p_{k})\\ \end{split} $ $ = A(1;1)A(1;2) + A(1;1, 2) - A(2;1, 2)$
For $J = 3$ and $K = 3$,
$\begin{split} f(\vec{x}) & = & \left (\sum_{k=1}^{3} c_{1k} x_{k} \right ) \left (\sum_{k=1}^{3} c_{2k} x_{k} \right) \left (\sum_{k=1}^{3} c_{3k} x_{k} \right) \\ & = & \sum_{k=1}^{3}\sum_{l=1}^{3}\sum_{m=1}^{3} c_{1k} x_{k} c_{2l} x_{l} c_{3m} x_{m} \end{split} $
$\begin{split} \sum_{\vec{x}} f(\vec{x}) P(\vec{X} = \vec{x}) & = &\sum_{x_{1} = 0}^{1}\sum_{x_{2} = 0}^{1} \sum_{x_{3} = 0}^{1} f(x_{1}, x_{2}, x_{3}) P(X_{1} = x_{1}, X_{2} = x_{2}, X_{3} = x_{3}) \\ &= &\sum_{x_{1} = 0}^{1}\sum_{x_{2} = 0}^{1} \sum_{x_{3} = 0}^{1} \sum_{k=1}^{3}\sum_{l=1}^{3}\sum_{m=1}^{3} c_{1k} x_{k} c_{2l} x_{l} c_{3m} x_{m} P(X_{1} = x_{1}, X_{2} = x_{2}, X_{3} = x_{3}) \\ &= & \sum_{k=1}^{3}\sum_{l=1}^{3}\sum_{m=1}^{3} I(k \ne l, l \ne m, m \ne k) c_{1k} c_{2l} c_{3m} P(X_{k} = 1, X_{l} = 1, X_{m} = 1) \\ &{}& + I(k = l, l \ne m) c_{1k} c_{2l} c_{3m} P(X_{l} = 1, X_{m} = 1) \\ &{}& + I(l = m, m \ne k) c_{1k} c_{2l} c_{3m} P(X_{m} = 1, X_{k} = 1) \\ &{}& + I(m = k, k \ne l) c_{1k} c_{2l} c_{3m} P(X_{k} = 1, X_{l} = 1) \\ &{}& + I(m = k = l) c_{1k} c_{2l} c_{3m} P(X_{k} = 1) \\ \end{split}$ $ = \left [A(1;1)A(1;2)A(1;3) - A(2;1,2)A(1;3) + A(3;1,2,3) - A(2;2,3)A(1;1) + A(3;1,2,3) - A(2;3,1)A(1;2) + A(3;1,2,3) - A(3;1,2,3) \right ] + A(1;1,2)A(1;3) - A(2;1,2,3) + A(1;2,3)A(1;1) - A(2;1,2,3) + A(1;3,1)A(1;2) - A(2;1,2,3) + A(1;1,2,3)$
$ = A(1;1)A(1;2)A(1;3) + [A(1;1,2) - A(2;1,2)]A(1;3) + [A(1;2,3) - A(2;2,3)]A(1;1) + [A(1;3,1)- A(2;3,1)]A(1;2) + A(1;1,2,3) - 3A(2;1,2,3) + 2A(3;1,2,3)$
For $J = 4$,
$ \lbrace A(1;1)A(1;2)A(1;3)A(1;4) - [A(2;1,2)A(1;3)A(1;4) - A(3;1,2,3)A(1;4) - A(3;1,2,4)A(1;3) + A(4;1,2,3,4)] ... \rbrace + [A(1;1,2)A(1;3)A(1;4) - A(2;1,2,3)A(1;4) - A(2;1,2,4)A(1;3) + A(3;1,2,3,4)] ... + [A(1;1,2,3)A(1;4) - A(2;1,2,3,4)] ... + A(1;1,2,3,4)$
$ =A(1;1)A(1;2)A(1;3)A(1;4) + [(A(1;1,2) - A(2;1,2))A(1;3)A(1;4) - (A(2;1,2,3)- A(3;1,2,3))A(1;4) - (A(2;1,2,4) - A(3;1,2,4))A(1;3) + A(3;1,2,3,4) - A(4;1,2,3,4)] ... + [A(1;1,2,3)A(1;4) - A(2;1,2,3,4)] ... + A(1;1,2,3,4)$