Show that the expected total present value of the bonds > purchased by time $t$ is $1000\lambda(1-e^{-rt})/r.$

Question

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 $$\frac{1000\lambda(1-e^{-r \ t})}{r}.$$

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

What am I missing?

Answer 1

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 $T_k\le t$, 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} E\left[\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}^\infty\frac{(\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 $$ E[\sum_{k=1}^\infty f(T_k)]=\lambda\int_0^\infty f(x)\,dx $$

Answer 2

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 $$\begin{align} E[V_t] &= PE\left[\sum_{i=1}^{I_t} e^{-rT_i}\right] \\ &= P\sum_{k=1}^{\infty}E\left[\sum_{i=1}^{k} e^{-rT_i}\Bigg|I_t = k\right]\Pr\{I_t=k\} \\ &= P\sum_{k=1}^{\infty}E\left[\sum_{i=1}^{k} e^{-rU_i}\Bigg|I_t = k\right]\Pr\{I_t=k\} \\ &= P\sum_{k=1}^{\infty}\sum_{i=1}^{k}E\left[ e^{-rU_i}\right]\Pr\{I_t=k\} \\ &= PE\left[ e^{-rU_1}\right]\sum_{k=1}^{\infty} k\Pr\{I_t=k\} \\ &= PE\left[ e^{-rU_1}\right] E[I_t]\\ \end{align}$$ 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

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