Can someone help me with the following annuity:
For $n \in \mathbb{N}, x \geq 0, v = \frac{1}{1+i}$ with $i$ being the interest rate, look at the following annuity, where the present value is given by: $$Y = v^{T_x} \overline{a}_{n-T_x} \mathbf{1}_{T_x \leq n}$$
where $T_x$ is the remaining life time of an $x$ year old Person with $P(T_x \leq t) = {}_t q_x$ and $P(T_x > t) = {}_t p_x$. . Note that for the index of $\overline{a}$ there should be an angle, since it is actuarial Notation.
The question now is to first of all describe, how one can Interpret this annuity and then calculate $E[Y]$ in dependence of $\overline{a}_n$ and $\overline{a}_{x:n}$.
Now $$\overline{a}_x = \frac{1-E[v^{T_x}]}{\delta} = \frac{1-\overline{A_x}}{\delta}$$
with $\delta = \log(1+i)$ being the force of intensity.
My thouhgt was, that I can rewrite the first part as follows:
$$ v^{T-x} \overline{a}_{n-T_x} = v^{T_x} \frac{1-v^{n-T_x}}{\delta} = \frac{1-v^n}{\delta} - \frac{1-v^{T_x}}{\delta} = \frac{v^{T_x}-v^n}{\delta}$$
And now? Just apply the expectation and then done? This would yield:
$$E[Y] = \frac{E[v^{T_x}] - v^n}{\delta} P(T_x \leq n) = \frac{\overline{A_x}-v^n}{\delta} {}_n q_x$$
but this doesn't seem to be right, does it?