Dtyjlui

Investors purchase $1000$ dollar bonds at the random times of a
Poisson process with parameter $lambda$. If the interest rate is
$r$, then the present value of an investment purchased at time $t$ is
$1000e^{-r t}$. Show that the expected total present value of the bonds
purchased by time $t$ is
$$frac{1000lambda(1-e^{-r t})}{r}.$$

I'm not sure how what kind of random variable the present value is here. But let $I_tsimtext{poi}(lambda t)$ be the number of purchases at time $t$, denote the present value by $V_t$, so we have that $V_t=1000cdot I_tcdot e^{-r t}.$ I doubt this is correct though, since this is now hard to find $E(V_t).$

What am I missing?

Here is another way to look at it. I will replace $1000$ with $1$ for readability.

Let $T_k$ be the time of the $k^{th}$ investment. Recall that the pdf $f_k$ of $T_k$ is a Gamma distribution: $$f_{k}(s)=lambda^k frac{s^{k-1}}{(k-1)!}e^{-lambda s}.$$ Furthermore, the present value of $T_k$ is equal to $e^{-rT_k}$ if $T_kle t$, and is zero if $T_k>t$ (since we only want to count investments before time $t$). Therefore, the expected total present value of all investments is
begin{align}
Eleft[sum_{k=1}^infty PV(T_k)right]
&=sum_{k=1}^infty E[PV(T_k)]
\&=sum_{k=1}^infty int_0^infty PV(s)f_k(s),ds
\&=sum_{k=1}^infty int_0^t e^{-rs} frac{lambda^k s^{k-1}}{(k-1)!}e^{-lambda s},ds
\&= int_0^t lambda e^{-rs} Big(sum_{k=1}^inftyfrac{(lambda s)^{k-1}}{(k-1)!}Big)e^{-lambda s},ds
\&= int_0^t lambda e^{-rs} e^{lambda s}e^{-lambda s},ds
\&= int_0^t lambda e^{-rs} ,ds
\&= lambda (1-e^{rt})/r
end{align}

In general, if $f:mathbb R^+to mathbb R$ is any integrable function, then
$$
E[sum_{k=1}^infty f(T_k)]=lambdaint_0^infty f(x),dx
$$

Let $T_i$ be the purchasing time of the $i$-th bond with value $P$. The total present value of the bond purchased up to time $t$ is

$$ V_t = sum_{i=1}^{I_t} Pe^{-rT_i}$$

with the convention that $V_t = 0$ when $I_t = 0$. Then the expected value is
$$begin{align}
E[V_t]
&= PEleft[sum_{i=1}^{I_t} e^{-rT_i}right] \
&= Psum_{k=1}^{infty}Eleft[sum_{i=1}^{k} e^{-rT_i}Bigg|I_t = kright]Pr{I_t=k} \
&= Psum_{k=1}^{infty}Eleft[sum_{i=1}^{k} e^{-rU_i}Bigg|I_t = kright]Pr{I_t=k} \
&= Psum_{k=1}^{infty}sum_{i=1}^{k}Eleft[ e^{-rU_i}right]Pr{I_t=k} \
&= PEleft[ e^{-rU_1}right]sum_{k=1}^{infty} kPr{I_t=k} \
&= PEleft[ e^{-rU_1}right] E[I_t]\
end{align}$$
where $U_i sim text{Uniform}(0, t)$ and they are independent. The second step is just applying the law of total expectation, and the third step is a key step: since the time $(T_1, T_2, ldots, T_{k})$ given $I_t = k$ are jointly follows ordered uniform on $0, t)$, and we are computing the expectation of a symmetric function of those time, we can replace it with the unordered uniform $U_i$, and they are i.i.d.. Therefore subsequently we can pull them out from the summation.

Next we compute
$$ E[e^{-rU_1}] = int_0^t e^{-ru}frac {1} {t}du = frac {1 - e^{-rt}} {rt}$$

As $I_t sim text{Poisson}(lambda t)$, $E[I_t] = lambda t$ and we conclude that

$$ E[V_t] = P times frac {1 - e^{-rt}} {rt} times lambda t
= frac {Plambda(1 - e^{-rt})} {r}$$