Given a discrete probability distribution (e.g., ${P_1=0.85,P_2=0.05,P_3=0.05,P_4=0.05}$), I would like to transform it according to some set of "weights" (say, ${w_1=2,w_2=0.5,w_3=1,w_4=0.5}$), which in this case would increase $P_1$ by some amount, decrease $P_2,P_4$ by some amount, and leave $P_3$ the same. Simply multiplying $p_i w_i$ won't do. I was thinking along the lines of casting both as a matrix, but the question would then be, what properties would W need to satisfy such that $\Sigma p_i$ is always 1?
Modifying a discrete probability distribution according to set of weights
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probability
terminology
transformation
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0Won't _any_ transformation do? If $Y = g(X)$ where $g$ is any (bounded) function, then for any $\beta$, $$P\{Y = \beta\} = \sum_{\alpha ~ \colon ~\beta = g(\alpha)} P\{X = \alpha\}.$$ Where do you think the probability mass will disappear to (or appear from) to make the sum of probabilities in the new distribution less than (or more than) $1$? – 2011-12-05
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0@Dilip Sarwate: Given a discrete probability distribution (e.g., $\{ P_1 = 0.85, P_2 = 0.05, P_3 = 0.05, P_4 = 0.05 \}$, I would like to transform it according to some set of "weights" (say, $\{ w_1 = 2, w_2 = 0.5, w_3 = 1, w_4 = 0.5 \}$), which in this case would increase $P_1$ by some amount, decrease $P_2,P_4$ by some amount, and leave $P_3$ the same. Simply multiplying $p_i w_i$ won't do. I was thinking along the lines of casting both as a matrix, but the question would then be, what properties would W need to satisfy such that $\Sigma p_i$ is always 1? – 2011-12-05
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1What you seemed to be asking and what you seem to want are very different things. Why not incorporate your above response into a revised question so people can understand what it is that you are asking, and respond appropriately? – 2011-12-05
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0@Dilip Sarwate: Done. I'll leave my comment, unless Stack Exchange protocol says to delete it. – 2011-12-05