# How to prove the variance of a function g(X) of the random variable X is given by $var[g(X)]\approx\{\frac{dg(X)}{dX}\}^2var(X)$

I am reading the book Modelling Survival Data in Medical Research. The book says that it's the Taylor series approximation to the variance of a function of a random variable. I learnt Taylor expansion back in high school and as far as I could recollect from my memory, Taylor expansion basically says that the function could be approximated by a polynomial:

If expanding f(x) at x=0, the Taylor expansion of f(x) would look like:

$$f(x)=f(0)+f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^2+...+o(1)$$

I don't know exactly what o(1) is because I never got a chance to know more advanced maths, but I know at the limit 0(1) should approach to 0.

However, I am still quite blind in deriving the approximation of variance in this case. Could someone please show me the step? I explain my understanding in Taylor series because I was hoping someone could point out if I have misunderstood the theories or if there are other relevant and important theorems and knowledge in deriving the variance approximation. Many thanks!

## 1 Answer

A first-order approximation should be enough. $$g(X) \approx g(\bar{X}) + \frac{d g(\bar{X})}{d X} (X-\bar{X})$$ Where $$\bar{X}$$ - some determined (fixed) value. Therefore, $$g(\bar{X}) - \frac{d g(\bar{X})}{d X} \bar{X}$$ is fixed term in the above expression and has zero variance.

From the property of variance $$\text{Var}\{aX\} = a^2 \text{Var}\{X\}$$ for any fixed value $$a$$ the desired expression follows immediately: $$\text{Var}\{g(X)\} \approx [\frac{d g(\bar{X})}{d X}]^2 \text{Var}\{X\}$$