# What is the expected present discounted value of a variable following a geometric Brownian motion?

According to Investment under Uncertainty by Dixit & Pindyck (1993, p.82), the expected present discounted value of a variable following a geometric Brownian motion is defined as: $$E\left[\int_{0}^{\infty}{F(x(t))}\exp[-rt]dt\right]$$ Suppose $$\tag 1 F(x)=x^\theta$$ and $$\tag 2 dx={α}{x}{dt}+{σ}{x}{dz}$$ where $$dz$$ is the increment of a Weiner process. Equation $$(1)$$ then follows a geometric Brownian motion. Exapanding equation $$(1)$$ by Itô's lemma, we can derive $$\tag 3 \theta{x}^\theta({α}{x}{dt}+{σ}{x}{dz})+\frac{1}{2}\theta(\theta-1){x}^\theta\sigma^2dt$$ Substituting equation $$(1)$$ for $${x}^\theta$$ and rearranging gives $$\tag 4 [\theta{α} + \frac{1}{2}\theta(\theta-1)\sigma^2]Fdt + \theta\sigma{F}dz$$ If $$x(0)$$ is currently $${x_0}^\theta$$, the expected value of $$F(x)$$ is given by equation $$(5)$$. $$\tag 5 {x_0}^\theta\exp[t(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2)]$$ According to the textbook mentioned above, the present discounted value of equation $$(5)$$ is given by equation $$(6)$$ $$\tag 6 \frac{x_0^\theta}{r-(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2)}$$

As far as I understand, deriving this result involves multiplying equation $$(5)$$ by $$\exp[-rt]$$ and integrating with respect to time between $$t=0$$ and $$t=\infty$$. Doing this, we arrive at the integral given by equation $$(7)$$ $$\tag 7 \int_{0}^{\infty}\frac{x_0^\theta}{\exp[t(r-(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2))]}dt$$ However, I do not understand how $$(5)$$ follows from $$(7)$$, if this is indeed the correct approach. Any assistance in understanding this problem would be appreciated.

The anti-derivative of (7) is $$\left[\frac{- x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t (r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right]^\infty_0$$ Now evaluate as $$t \rightarrow \infty$$ and $$t \rightarrow 0$$ $$\lim_{t \rightarrow \infty} \left[\frac{- x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t(r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right] + \lim_{t \rightarrow 0} \left[\frac{x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t(r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right]$$ $$0 + \frac{ x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}$$ Which is equal to equation (6).