Two observations:
- If $X \sim \text{Geom}(p)$ then $E(X) = \frac{1}{p}$ (using the "first success" definition, i.e., X is supported on the set $\{1, 2, 3, ...\}$).
- The present value of a perpetuity, $\text{PV}$, is the constant cashflow, $\text{CF}$, multiplied by this expectation (when the discount rate is $p$): $\text{PV} = \frac{\text{CF}}{p}$.
I'm trying to connect these concepts in an intuitive way and try to explain what this expectation is in the context of a never-ending stream of cashflows. Two questions:
- What is the random variable $X$ in the context of a perpetuity?
- How is $E(X)$ interpreted (as an average)?
I want to fill in the blank for the the statement:
A perpetuity is a never-ending stream of future, periodic cashflows (each cashflow in the amount of $\text{CF}$) discounted at the rate, $p$. It can be calculated by multiplying the cashflow amount by the average _____________.
There is no connection in the two formulas.
$$\mathbb{E}[X]=\Sigma_{x=1}^{\infty}x\cdot p(1-p)^{x-1}=\dots=\frac{1}{p}$$
is the result of the definition of the expectation of a discrete random variable
$$\mathbb{E}[X]=\Sigma_x x\cdot p(x)$$
Where
$p$ is a probability: the probability of getting a success
$p(1-p)^{x-1}$ is a Probability Mass Function
The proof of the result of these sum is quite simple and I leave you it as an exercise
The present value of a perpetuity is the sum of a deterministic, periodic and constant amount $CF$ shifted to the present value using a fixed interest rate $p$. Just to avoid misunderstanding let's set the interest rate as $i$
Thus the PV is
$$PV=CF[(1+i)^{-1}+(1+i)^{-2}+\dots +]$$
Immediately, setting $(1+i)^{-1}=c$ you get
$$PV=CF[c+c^2+c^3+\dots +]=\frac{CF c}{1-c}=\frac{CF}{1+i}\frac{1}{1-\frac{1}{1+i}}=\frac{CF}{i}$$