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Let $t_1,\ldots,t_m$ be $m$ random variables that are independently and identically drawn from a Bernoulli distribution with a constant parameter $p$.

Now, we define some functions of $t_1,\ldots,t_m$, for constant $w_1,\ldots,w_m$, in the following way:

$$f_i(t_1,\ldots,t_m)=t_i\mathbf{I}(1-\sum_{j=1}^m w_jt_j\geq 0)$$

where $\mathbf{I}(x\geq 0)=1$ if $x\geq 0$, and $\mathbf{I}(x\geq 0)=0$ if $x< 0$.

My question: what is the joint expected value of $f_i$'s as a function of $w_1,\ldots,w_m$ and $p$.

$$g_i(w_1,\ldots,w_m,p)=\mathbf{E}[f_i(t_1,\ldots,t_m)]=?$$

Edit:

If $f(t_1,\ldots,t_m)=[f_1(t_1,\ldots,t_m),\ldots,f_m(t_1,\ldots,t_m)]^T$, then I am interested in:

$$g(w_1,\ldots,w_m,p)=\mathbf{E}_{t_1,\ldots,t_m~\text{iid}\sim Bernoulli(p)}[f(t_1,\ldots,t_m)]$$

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1 Answer 1

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${\bf E} [f_i(t_1,\ldots,t_m)]$ is the probability that $t_i = 1$ and $\sum_j w_j t_j \le 1$. Conditioning on $t_i$, we get $${\bf E} [f_i(t_1,\ldots,t_m)] = p\; {\bf P}\left\{ \sum_{j \ne i} w_j t_j \le 1 - w_i\right\}$$ I doubt that there's a simpler expression for this. Even determining whether ${\bf P}\{\sum_j w_j t_j =0\} > 0$ (where the $w_j$ are integers), i.e. whether there is a subset of the $w$'s whose sum is $0$, is an NP-complete problem (Subset sum problem).

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    $\begingroup$ The expected value of the vector is the vector of expected values of the entries. $\endgroup$ Oct 29, 2014 at 18:21
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    $\begingroup$ Can't we calculate ${\bf P}\left\{ \sum_{j \ne i} w_j t_j \le 1 - w_i\right\}$ by a convolution-based method? $\endgroup$
    – Alt
    Oct 29, 2014 at 19:13
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    $\begingroup$ Unless I misunderstand, essentially your "convolution-based method" will be calculating the probabilities for all the values of all the partial sums. The problem is that in general there will be $2^m$ different possible values to consider. $\endgroup$ Oct 29, 2014 at 20:20

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