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I'm a bit stuck on the following question: Suppose $X$ is a Geometric Brownian Motion with a drift $\mu$ and volatility $\sigma$. How can you expand, or approximate, the function $f(X)=(a + bX)^c$ in terms of powers of $X$, where $a$, $b$, and $c$ are real numbers? Can anyone solve it?

Eastwood94
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    Expand or approximate leaves a lot of room for interpretation. You should be able to "approximate" it by a Taylor polynomial of order $\lfloor c\rfloor,.$ – Kurt G. Nov 12 '23 at 11:54
  • Thank you. That was my first attempt, but taking the expected value of f(X), which by linearity reduces to a sum including all the moments of X, I get that the result diverges. Any idea how to solve it? – Eastwood94 Nov 12 '23 at 14:09
  • Let's write the GBM as $$X_t=X_0e^{\sigma W_t+\mu t-\frac{\sigma^2 t}{2}},.$$ As we know the $n$-th moment of this is $$\mathbb E[X^n_t]=X_0^ne^{n\mu t+(n^2-n)\frac{\sigma^2 t}{2}},.$$ This does not diverge, and neither does a truncated Taylor expansion up to order $n=\lfloor c\rfloor,.$ The fun probably starts when you want an infinite Taylor series. I have (lazy as I am) not worked on this since your question left room for interpretation. Where does this come from? What do you want to achieve? – Kurt G. Nov 12 '23 at 15:17
  • That’s the expression for the n-th moment I am using. That diverges as n tends to $\infty$, even when dividing it by $n!$, such as for obtaining the expectation $E(e^{(X)})=\sum_{n=0}^{\infty}\frac{1}{n!}E(X^n)$ – Eastwood94 Nov 12 '23 at 18:47
  • But what has $e^X$ to do with $f(X)=(a+bX)^c,?$ – Kurt G. Nov 12 '23 at 19:12
  • I just wanted to point out that I don't understand the reason for these divergences, and to show that the same thing happens when applied to a function such as the exponential. Returning to the function of my original question, the expected value would be: $E[f(X)]= a^c \cdot \sum_{n=0}^{\infty} \pmatrix{c \n} b^{n}a^{-n}E[X^n] = a^c \cdot \sum_{n=0}^{\infty} \pmatrix{c \n} b^{n}a^{-n} X_0^n e^{n\mu t + \frac{n(n-1)}{2}\sigma^2 t}$ , and the n-th coefficient of this sum tends to infinity as n increases. – Eastwood94 Nov 12 '23 at 19:36
  • It is well-known that for a GBM $X$ the expectation $\mathbb E[e^X]$ is infinite. Not everything can be finite. If you are not just playing around you may want to let us know what the purpose of $f(X)=(a+bX)^c$ is and what your typical $c,$s look like. – Kurt G. Nov 12 '23 at 19:38
  • I was hoping to get or approximate the expectation of $\mathbb{E}[(a+bX)^c]$, where $c\in(0,1]$, so I asked for the expansion in terms of powers of $X$, since that was my first intuition. Any ideas? – Eastwood94 Nov 13 '23 at 10:37
  • It was highest time to mention what values that $c$ takes. I am getting tired of this question. Hint: study what the best way is to Taylor expand the sqrt function. – Kurt G. Nov 13 '23 at 11:14

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I suggest that you use Taylor's expansion with binomial series:

$$f(X)=a^c\left(1+\frac{b}{a}X\right)^c = a^c \cdot \sum_{n=0}^{\infty} \left(\pmatrix{c \\n} b^{n}a^{-n}X^n \right)$$ where $$\pmatrix{c \\n}=\frac{c(c-1)...(c-n+1)}{n!}$$

NN2
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  • Thank you. That was my first attempt, but taking the expected value of $f(X)$, which by linearity reduces to a sum including all the moments of $X$, I get that the result diverges. Any idea how to solve it? – Eastwood94 Nov 12 '23 at 13:54
  • @Eastwood94 I think the series must converges. Could you give me the expression of $\mathbb{E}(X^n)$ that you computed? – NN2 Nov 12 '23 at 14:26
  • The expression I use is: $E[X^n] = X_0^n e^{n\mu t + \frac{n(n-1)}{2}\sigma^2 t}$ – Eastwood94 Nov 12 '23 at 19:37
  • @Eastwood94 I thought $X$ followed the Brownian motion. But if $X$ follows the geometric brownian motion, the series does not converge. Indeed, except some specific values of $(a,b,c)$, $\mathbb{E}(f((a+bX)^c) = \infty$ in general. – NN2 Nov 13 '23 at 12:20
  • Thanks for your clarification @NN2 – Eastwood94 Nov 13 '23 at 12:32