Proving that weighted probabilities lie in the range 0 to 1

How to prove that:

$$P_{avg} = \dfrac{w_{1}P_{A} + w_{2}P_{B} + w_{3}P_{C}}{w_{1} + w_{2} + w_{3}}$$

lies between 0 and 1, where $P_{i}$ corresponds to probability scores and $w_{j}$ are real numbers

• The conclusion cannot be proven unless you assume the $w_j$ are nonnegative real numbers. – whuber Jan 17 '16 at 21:06
• The answers and the comment by @whuber are true, under the presumption that all the probabilities lie in the range 0 to 1. As I have shown at stats.stackexchange.com/questions/4220/… , this presumption is not always true in the "real world". – Mark L. Stone Aug 9 '16 at 22:01

It is the same way in which one proves that any weighted average lies between its maximum and minimum component. Consider the set $$A=(P_{A},P_{B},P_{C})$$ Without loss of generality, let ${max{A}=P_{A}}$and ${min{A}=P_{B}}$

Now, consider weights $$0\leq\omega_{i}\leq1$$ such that $$\sum_{i}\omega_{i}=1$$ . Therefore, $$0\leq P_{B}\leq\tilde{P}=\sum_{i}\omega_{i}P_{i}\leq P_{A}\leq1$$

• I am not sure, if I understand the part in the equality where you put weighted sum between min and max. I understand intuitively that if you replace all probabilities by max probability and then take a weighted sum, it will be equal to max. As all other probabilities are less than max, hence the score will be less.. – The Wanderer Jan 17 '16 at 20:45
• Your intuition is correct- the max probability with weight 1 will be weakly higher than a sum of weighted probabilities which contain elements (weakly) lesser than the max probability. – ChinG Jan 17 '16 at 20:50

There are probably a lot of ways to prove this, but here's my shot.

Let $\tilde{w_i} = w_i/(w_1+w_2+w_3)$, $W = (\tilde{w_1}, \tilde{w_2}, \tilde{w_3})'$, $P = (P_A,P_B,P_C)'$ and $|| \cdot ||_2$ denote the $L2$ norm. Then $$P_{avg} = |W'P| \le ||W||_2 |||P||_2 \le 1$$

by the Cauchy-Schwarz inequality.

• (1) This does not appear to be an accurate statement of the Cauchy-Schwarz Inequality, unless you mean some unusual things by "$||\ ||$". (2) You still need to prove that $||W||$ and $||P||$ do not exceed $1$. Why not just use Holder's Inequality? – whuber Jan 17 '16 at 21:16
• 1) you're right; I changed the notation a bit. 2) That's pretty much by assumption. CS is just a special case of Holder. – Taylor Jan 17 '16 at 21:36