# Modifying a discrete probability distribution according to set of weights

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?

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Won'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) $1$? –  Dilip Sarwate Dec 5 '11 at 13:49
@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? –  higgy Dec 5 '11 at 15:21
What 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? –  Dilip Sarwate Dec 5 '11 at 15:34
@Dilip Sarwate: Done. I'll leave my comment, unless Stack Exchange protocol says to delete it. –  higgy Dec 5 '11 at 15:40

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.

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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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This 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? –  higgy Dec 5 '11 at 16:50
You could say that $\sum w_i p_i=1$, but I don't see any other way to specify the requirement. –  Ross Millikan Dec 5 '11 at 17:00