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We are given a finite collection of random variables, $X_1,\ldots,X_n$. Each can take exactly $k$ different real values, with the same probability, $1/k$, but a priori these values are not the same for each of the variables. Thus, if $X_i$ can assume values $x_1^i,\ldots,x_k^i$, its probability distribution is given by: $$\mathbb{P}[X_i=x]=\begin{cases}1/k\quad if\ x\in {x_1^i,\ldots,x_k^i}\\ 0\quad otherwise \end{cases}$$ We now take a linear combination of the random variables, and define $$ Y=\sum_i a_iX_i,\quad a_i>0, s.t. \sum_i a_i =1$$ How can we describe the probability distribution of this new random variable? My final goal would be, given $k\in\mathbb{R}$, determining the right coefficients to maximise $$\mathbb{P}[Y>k]$$

What I would like to do is to write the new probability distribution of the the random variable $Y$, that of course will be dependant on the coefficients $\left\{a_i\right\}_i$, and if possible its associated cumulative distribution. So far I haven't been able to get to any "closed form" for the probability distribution of "Y". My hunch would be to say that "Y" still has an equally distributed probability function, in the generic case, and that I should therefore choose the coefficients $a_i$ in such a way that its support is "as shifted as possible" so that it would be greater than $k$ as much as possible. Am I on the right path? Any help would be very appreciated. Thank you!

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    $\begingroup$ You should add a constraint on coefficients $a_i$ such as having a sum equal to 1. Otherwise it suffices to take say all $a_i$ eqal to $10^{100}$ in order to have \mathbb{P}[Y>k] very very close to 1... $\endgroup$ – Jean Marie Aug 1 '16 at 21:39
  • $\begingroup$ Thank you for the comment, I'll edit the question. Nevertheless, I don't get why you would get such a high probability. What if $X_j^k$ are all negative and the problem admits no solution, for instance? $\endgroup$ – fatoddsun Aug 1 '16 at 21:47
  • $\begingroup$ 1) It's true that I had assumed that the $X_i$s were taking $\geq 0$ values. 2) Why don't you consider using generating functions ? If $S_{n}=\sum _{i=1}^{n}a_{i}X_{i}$ then $G_{S_{n}}(z)=\operatorname {E} (z^{S_{n}})=\operatorname {E} (z^{\sum _{i=1}^{n}a_{i}X_{i},})=G_{X_{1}}(z^{a_{1}})G_{X_{2}}(z^{a_{2}})\cdots G_{X_{n}}(z^{a_{n}}).$ (borrowed to (en.wikipedia.org/wiki/Probability-generating_function)). $\endgroup$ – Jean Marie Aug 1 '16 at 22:07

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