For a problem I am working on I need to compute the conditional expectation of a continuous random variable $T$ given a discrete random variable $K$. I already derived a formula for the joint distribution function $P(T\leq t, K = k)$, as well as the pdf of $T$ and the probability mass function of $K$. Is there some formula to compute $\mathbb{E}\left[T|K=k \right] $? (And just out of interest, is there a formula for $\mathbb{E}\left[K|T=t \right] $?)
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$\begingroup$ From $P(T\leq t, K = k)$ one can derive $P(T\leq t|K = k)$, which is the conditional cdf of $T$ given $K$. From there one can either derive the conditional density or directly compute the expectation from the cdf. $\endgroup$– Xi'anCommented May 21, 2023 at 9:27
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The question of how to compute $E[T|K = k]$ has been addressed by Xi'an in the comment. More explicitly, assume $p_k := P(K = k) > 0$ for each $k$, then the conditional density of $T$ given $T = k$ is $f(t|K = k) = \frac{f(t, k)}{p_k}$, where $f(t, k) = \frac{dF_k(t)}{dt}$ and $F_k(t) = P[T \leq t, K = k]$. It thus follows that \begin{align} E[T|K = k] = \int_\mathbb{R} tf(t|K = k)dt = p_k^{-1}\int_\mathbb{R}tf(t, k)dt. \end{align}
Your second question about determining $E[K|T = t]$ is much more technical and a complete answer requires measure-theoretic machinery. Compared with the first question, the difficulty lies in that the conditional distribution of $K$ given $T = t$ can no longer be evaluated by the elementary event-based conditional probability formula $P(A | B) = P(A \cap B)/P(B)$ as $P[T = t] = 0$ when $T$ is continuous. To begin with, it is OK to expand $E[K | T = t]$ as \begin{align} E[K | T = t] = \sum_k kP(K = k|T = t). \tag{1} \end{align}
Coming to determine $P(K = k|T = t)$, because of the aforementioned technicality, it should be understood as the a.s. value of the measure-theoretic conditional probability (which is a random variable of $T$) $g(T) := P(K = k|T) = P(K = k|\sigma(T))$ on the set $[T = t]$. Exercise 33.16 in Probability and Measure (3rd edition) by Patrick Billingsley connected this value with the joint distribution function $P[T \leq t, K = k]$: \begin{align} & P(K = k | T = t) = \lim_{h \to 0}P[K = k | t - h < T \leq t + h] \\ =& \lim_{h \to 0} \frac{P(T \leq t + h, K = k) - P(T \leq t - h, K = k)}{P(t - h < T \leq t + h)}. \tag{2} \end{align}
With the available information you stated, $(1)$ and $(2)$ together should be sufficient to determine $E[K | T = t]$.
