# Prove that deferred continuous annuity has APV: $E(e^{-\int_w^t \delta(s)ds}\int_0^{T-w}e^{-\int_w^{w+t} \delta(s)ds}dt\mathbb I_{T>w})$

I want to prove that deferred continuous annuity has the APV:

$$E(e^{-\int_w^t \delta(s)ds}\int_0^{T-w}e^{-\int_w^{w+t} \delta(s)ds}dt\mathbb I_{T>w})$$

$$=E(\int_0^{T-w}e^{-\int_0^{w+t} \delta(s)ds}dt\mathbb I_{T>w})$$

Where $$T$$ is the lifetime RV, $$w$$ is the point at which the annuity starts $$\delta$$ is the interest.

The formula is quite obvious, but how do I prove it? What I need to show is that it equals:

$$\text{lim}_{m\rightarrow \infty }\frac{1}{m}\sum_{k=0}^\infty e^{-\int_0^{w+k/m} \delta(s)+\mu(x+m)ds}$$

$$=\text{lim}_{m\rightarrow \infty }\frac{1}{m}\sum_{k=0}^\infty e^{-\int_0^{w+k/m} \delta(s)ds }P(T>w+k/m)$$

$$=\text{lim}_{m\rightarrow \infty }\frac{1}{m}\sum_{k=0}^\infty e^{-\int_0^{w+k/m} \delta(s)ds }E(\mathbb I_{T>w+k/m})$$

Or at least equals: $$\int_w^{\infty}e^{-\int_0^{t} \delta(s)ds}\mathbb P(T>t)dt$$

The issue is, how do I take the expectation with regards to the integral?