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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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
2 Answers
It is not clear to me what exactly the OP wants.
Let us call a $n$-tuple $(x_1, x_2, \cdots, x_n)$ a stochastic vector if $0 \leq x_i \leq 1$ for $1 \leq i \leq n$, and $\sum_{i=1}^n x_i = 1$. Let us call a stochastic vector a proper stochastic vector if each $x_i > 0$. We are given a proper stochastic vector $\mathbf P = (P_1, P_2, \cdots, P_n)$. Each real vector $\mathbf w = (w_1, w_2, \cdots, w_n)$ defines a transformation $$w: \mathbf P \to w(\mathbf P) = (w_1P_1, w_2P_2, \cdots, w_nP_n).$$
Given an arbitrary stochastic vector $\mathbf x = (x_1, x_2, \cdots, x_n)$, is there a $\mathbf w$ such that $w(\mathbf P) = \mathbf x$?
Obviously yes. We have $w_i = x_i/P_i$ for $1 \leq i \leq n$. Note that the $w_i$ are all nonnegative real numbers.
Characterize the set of all $\mathbf w$ such that $w(\mathbf P)$ is a stochastic vector.
This is a lot harder. Obviously, it is necessary that each $w_i$ satisfies $0 \leq w_i \leq P_i^{-1}$, but as Ross Millikan says, there is not much else that can be said except that the $w_i$ must be such that $\sum w_iP_i = 1$, that is, $w(\mathbf P)$ must be a stochastic vector. We could dress it up probabilistically and say that the $\mathbf w$ are the set of all possible ranges that a nonnegative discrete random variable $X$ with $E[X] = 1$ can have, where the probability mass function of $X$ is constrained to be
$$p_X(w_i) = P\{X = w_i\} = P_i, ~ 1 \leq i \leq n$$ but where is the fun in that?
Most important, the transformation that the OP seeks has nothing to do with probability theory per se as I mistakenly thought in my initial comment on the question: we are not transforming one random variable into another and deriving the probability mass function of the image from the probability mass function of the source.
You need to rescale by the sum of the weights $W$. It sounds like you want $P_1:P_2:P_3P_4=1.7:0.025:0.05:0.025$, but you are correct the sum must be $1$. So $W=1.7+0.025+0.05+0.025=1.8$ and the new values are $P_1=1.7/1.8, P_2=0.025/1.8$, etc.
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0This makes sense, but what if I would like to avoid rescaling. For a probability distribution $P = \{ p_i \}$, I know that every $p_i$ must satisfy $(0 \leq p_i \leq 1)$. Above, I chose the weights $w_i$ arbitrarily, but could I place constraints on them such that when they are combined with the $p_i$ that the sum remains 1? – 2011-12-05
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0You could say that $\sum w_i p_i=1$, but I don't see any other way to specify the requirement. – 2011-12-05