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Let $P_1,...,P_k$ be probability measures on $(\Omega, \mathcal{F})$, and let $E \in \mathcal{F}$ with $P_i(E) > 0$, $i=1...k$. Suppose $\beta_i \geq 0$, $i=1,...,k$, and $\sum_{i=1}^k \beta_i = 1$.

Then, $\sum_{i=1}^k \beta_i P_i(\cdot \mid E)$ defines a probability measure on $(\Omega, \mathcal{F})$, because each conditional probability $P_i(\cdot \mid E)$ is a probability measure and a convex combination of probabilities is a probability.

My question is whether $\sum_{i=1}^k \beta_i P_i(\cdot \mid E)$ can be expressed as the conditional probability, given $E$, of a convex combination of $P_1,...,P_k$.

Question. Does there exist $\alpha_i \geq 0$, $i=1,...,k$, with $\sum_{i=1}^k \alpha_i = 1$ such that $(\sum_{i=1}^k \alpha_i P_i)(\cdot \mid E) = \sum_{i=1}^k \beta_i P_i(\cdot \mid E)$?

I believe the answer is yes because I think I can verify the case $i=2$ by a brute force calculation that I won't reproduce here (it's quite messy and, I think, not very instructive). The trouble is, I get the feeling that I'm missing some basic fact or observation that would answer my question in a more elegant and illuminating way.

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    It is clear now. – 2017-01-04

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Yes, this is really easy. Set $\alpha_i = \frac{\beta_i}{\alpha P_i(E)}$ with $\alpha = \sum_{i=1}^k\frac{\beta_i}{P_i(E)}$. Then $$ \left(\sum_{i=1}^k \alpha_i P_i\right)(\cdot \mid E) = \frac{\sum_{i=1}^k\alpha_i P_i(\cdot \cap E)}{\sum_{i=1}^k \alpha_i P_i(E)} =\frac{\sum_{i=1}^k \beta_i \frac{ P_i(\cdot \cap E)}{\alpha P_i(E)}}{ \sum_{i=1}^k \frac{\beta_i}{\alpha}} = \sum_{i=1}^k \beta_i P_i(\cdot \mid E). $$