I have two points (x1,y1 and x2,y2) that represent points in an exponential decay curve (discounted cash flows):
Exponential Decay using varying Discount Rates
The limits of my mathematics is using the linear slope/intercept equation to plot a line through these.
My (hopefully) simple question is: What is the equivalent exponential decay slope/intercept equation that I could use to plot a curve through these two points?
My aim is to determine the value of y when x = 0.
$\endgroup$ 21 Answer
$\begingroup$Assuming a relationship $$ y(x) = y_0 e^{-\lambda x} \quad (*) $$ If we take the logarithm of both sides, we get $$ z(x) = \ln(y) = \ln(y_0) -\lambda x \quad (**) $$ which is a linear curve.
From $(**)$ one can infer $y_0 = y(0)$ and $\lambda$ and use it to plot a graph of $(*)$.
Given two points $(x_1, y_1)$ and $(x_2, y_2)$, where $x_1 \le x_2$ the line $(**)$ is: \begin{align} z(x) &= (1-t) \ln(y_1) + t \ln(y_2) \quad (t \in [0,1]) \\ &= \left(1 - \frac{x-x_1}{x_2 - x_1}\right) \ln(y_1) + \frac{x-x_1}{x_2 - x_1} \ln(y_2) \\ &= \frac{x_2-x}{x_2 - x_1} \ln(y_1) + \frac{x-x_1}{x_2 - x_1} \ln(y_2) \\ &= \frac{(x_2-x)\ln(y_1)+(x-x_1)\ln(y_2)}{x_2 - x_1} \end{align} Then $$ \ln(y_0) = z(0) = \frac{x_2\ln(y_1)-x_1\ln(y_2)}{x_2 - x_1} $$ and $$ -\lambda = z'(x) = \frac{\ln(y_2)-\ln(y_1)}{x_2-x_1} $$Result:\begin{align} y_0 &= \exp\left(\frac{x_2\ln(y_1)-x_1\ln(y_2)}{x_2 - x_1}\right) \\ \lambda &= \frac{\ln(y_1)-\ln(y_2)}{x_2-x_1} \end{align} This gives the exponential curve: \begin{align} y(x) &= \exp\left(\frac{x_2\ln(y_1)-x_1\ln(y_2)}{x_2 - x_1}\right) \exp\left(-\frac{\ln(y_1)-\ln(y_2)}{x_2-x_1}\,x \right) \\ &= \exp\left( \frac{x_2\ln(y_1)-x_1\ln(y_2)}{x_2 - x_1} -\frac{\ln(y_1)-\ln(y_2)}{x_2-x_1}\,x \right) \end{align}
Example:
This resulted in $y_0 = 4.33$ and $\lambda = 0.37$.
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