A sequence of random variables, (not necessarily independent) are generated as follows. At each time $i \in [n]$, we toss a coin with head probability $\delta_i$ (these are random variables themselves). If the coin lands a head, we pay a cost $h_i$ and otherwise we pay a cost $t$. What is the expected cost paid before one of the coins lands tails? Specifically, how to deal with the fact that we stop after a tail lands?
Here is what I have tried, let $X_i$ be a r.v. which denotes the total cost paid till the $i^{th}$ iteration. Then, $\mathbb{E}[X_i] = \mathbb{E}[X_{i-1}+\delta_i]\delta_i+\mathbb{E}[X_{i-1}+t](1-\delta_i).$
However I am not sure how to use this fact. Can someone point me to some relevant literature for this problem?
Assuming independence, as otherwise it is unlikely the problem is tractable. Let $X_n=1$ if the $n^{\mathrm{th}}$ flip is heads and be zero otherwise. $X_n$ has a Poisson binomial distribution. Set $\tau = \inf\{n>0: X_n=0\}$. Then for each positive integer $n$, we have \begin{align} \mathbb P(\tau = n) &= \mathbb P(X_1=X_2=\cdots=X_{n-1}=1,X_n=0\}\\ &= \prod_{i=1}^{n-1} \mathbb P(X_i=1) \mathbb P(X_n=0)\\ &= \prod_{i=1}^{n-1} \delta_n (1-\delta_n). \end{align} Let $C_n$ denote the cost paid on the $n^{\mathrm{th}}$ flip. Since $\tau$ is a stopping time, it follows from Wald's identity that the expected cost paid before flipping a tails is given by \begin{align} \mathbb E\left[\sum_{i=1}^{\tau-1} h_iX_i\right] &= \sum_{i=1}^{\tau-1}h_i\delta_i(1-\delta_n). \end{align} We cannot say much more without further assumptions on the $h_i$ and $d_i$.