Suppose I have some quantity $L$ which can be $0$ with probability $(1 - \theta)$ or can be distributed according to a lognormal $\mathcal{LN}\left(\mu, \sigma^2\right)$ with probability $\theta$. That is:
$$\begin{aligned} L &= CR \\ C &\sim \text B\left(\theta\right) \\ R &\sim \mathcal{LN}\left(\mu, \sigma^2\right) \end{aligned}$$
Or equivalently:
$$\begin{aligned} P(C = 1) &= \theta \\ P(C = 0) &= 1 - \theta \\ \forall a, b > a: P(a < R < b) &= \frac {1}{\sigma\sqrt{2\pi}} \int\limits_{\max(0, a)}^{\max(0, b)} \frac 1 x \exp \left(-\frac {\left(\ln(x) - \mu \right)^2}{2 \sigma^2} \right) \text dx \\ P(L = 0|C = 0) &= 1 \\ P(L \neq 0 | C = 0) &= 0 \\ \forall a, b > a: P(a < L < b|C = 1, a < R < b) &= 1 \\ \forall a, b > a: P(a < L < b|C = 1, R < a \lor R > b) &= 0 \end{aligned}$$
What is $\mathbb E\left[L\right]$? Intuitively, it seems to be like it should be $\mathbb E\left[C\right]\mathbb E\left[R\right] = \theta \exp\left(\mu + \frac {\sigma^2}{2}\right)$, but I'm not sure how I'd prove that.
So far I've tried something like:
$$\begin{aligned} \forall a, b > a: P(a < L < b) &= \sum\limits_{c=0}^1 P(a < L < b|C=c)p(C=c) \\ &= (1 - \theta)\int\limits_a^b \delta(x) \text dx + \theta P(a < L < b|C=1) \\ &= (1 - \theta)\int\limits_a^b \delta(x) \text dx + \theta \int \limits_{-\infty}^{+\infty} P(a < L < b|R = r, C = 1) p(R = r|C = 1) \text dr \\ \end{aligned}$$
But I don't know if that's right or where to go from here.
In addition to the answer below, I also thought some more about the above way of describing things, and we can express the probability function conditional on $R = r$ as:
$$\forall a, b > a: P(a < L < b|C = 1, R = r) = \mathbf 1_{(a, b)} (r)$$
Therefore:
$$\begin{aligned} \forall a, b > a: P(a < L < b) &= (1 - \theta)\int\limits_a^b \delta(x) \text dx + \theta \int \limits_{-\infty}^{+\infty} P(a < L < b|R = r, C = 1) p(R = r|C = 1) \text dr \\ &= (1 - \theta)\int\limits_a^b \delta(x) \text dx + \theta \int \limits_{-\infty}^{+\infty} \mathbf 1_{(a, b)}(r) \frac {1}{\sigma\sqrt{2\pi}} \frac 1 r \exp \left(-\frac {\left(\ln(r) - \mu \right)^2}{2 \sigma^2} \right) \text dr \\ &= (1 - \theta)\int\limits_a^b \delta(x) \text dx + \theta \frac {1}{\sigma\sqrt{2\pi}} \int \limits_{max(0, a)}^{max(0, b)} \frac 1 r \exp \left(-\frac {\left(\ln(r) - \mu \right)^2}{2 \sigma^2} \right) \text dr \\ \end{aligned}$$
So $L$'s pdf is
$$p(L = l) = (1 - \theta) \delta (l) + \theta \frac {1}{\sigma\sqrt{2\pi}} \frac 1 l \exp \left(-\frac {\left(\ln(l) - \mu \right)^2}{2 \sigma^2} \right)$$
And its expected value is, then:
$$\begin{aligned} \mathbb E[L] &= (1 - \theta) 0 + \theta \exp \left(\mu + \frac{\sigma^2}{2} \right) \\ &= \theta \exp \left(\mu + \frac{\sigma^2}{2} \right) \end{aligned}$$
$$\operatorname{E}[L] = \operatorname{E}[CR] = \operatorname{E}[C]\operatorname{E}[R].$$ The right-hand side equality holds whenever $C$ and $R$ are independent.
Alternatively, $$\operatorname{E}[L] = \operatorname{E}[\operatorname{E}[L \mid C]]$$ by the law of total expectation. But $$\operatorname{E}[L \mid C] = \begin{cases} 0, & C = 0 \\ \operatorname{E}[R], & C = 1. \end{cases}$$ Therefore $$\operatorname{E}[L] = 0 \cdot \Pr[C = 0] + \operatorname{E}[R] \Pr[C = 1] = \theta \operatorname{E}[R].$$