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Let $V,T$ be two random variables with supports $\mathcal{V},\mathcal{T}$, respectively. Let

  • $P_{V|T}$ denote the probability distribution of $V$ coditional on $T$

  • $P_{V,T}$ denote the probability distribution of $(V,T)$

  • $P_{T}$ denote the probability distribution of $T$, $P_{V}$ denote the probability distribution of $V$

  • $P_{T|V}$ denote the probability distribution of $T$ coditional on $V$

Suppose that $\mathcal{V}$ and $\mathcal{T}$ are finite. Then, $$ E(V|T=t)= \sum_{v\in \mathcal{V}} v P_{V|T}(v|t)= \sum_{v\in \mathcal{V}} v \frac{P_{V,T}(v,t)}{P_{T}(t)}=\sum_{v\in \mathcal{V}} v \frac{P_{T|V}(t|v)P_V(v)}{P_{T}(t)} $$ Moreover, $$ \underbrace{E(V|T=t)\geq 0 \leftrightarrow \sum_{v\in \mathcal{V}} v P_{T|V}(t|v)P_V(v)\geq 0}_{(*)} $$ Now, I want to write condition $(*)$ (and in particular the right hand side) when $V$ and $T$ are continuous random variables, without using probability density functions but only probability measures or cumulative distribution functions. Could you help?

My attempt: I managed to write $(*)$ when $\mathcal{T}$ is finite and $V$ is a continuous random variable: $$ E(V|T=t)\geq 0 \leftrightarrow \int_{\mathcal{V}} v P_{T|V}(t|v)dP_V(v)\geq 0 $$ But I'm struggling to extend to the case when also $T$ is a continuous random variable.

TEX
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  • $P_V$ denotes the PDF, but in the integral you write it as if it is CDF, am I wrong? – gunes Jun 27 '19 at 11:47
  • In an answer (to a different but related question) at https://stats.stackexchange.com/questions/413331/, I provide an explicit, universal formula that answers your question in terms of the CDF. Elsewhere you can also find answers in terms of characteristic functions, moment-generating functions, and cumulant-generating functions. – whuber Jun 27 '19 at 12:11
  • $P_V$ is probability distribution and not density. – TEX Jun 27 '19 at 12:36

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