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I have a function which I would like to use Taylor expansion and calculate its variance by the following formula:

The formula for variance then becomes \begin{align} \operatorname{Var}(f(X))=[f'(EX)]^2\operatorname{Var}(X)+\frac{[f''(EX)]^2}{4}\operatorname{Var}^2(X)+\tilde{T}_3 \end{align}

got the formula from Variance of a function of one random variable

I have tried a few times to calculate a simple example to get to learn how to use it, but can not say I have succeeded.

So if anyone please could show me how to use the above formula for calculating the variance for this (simple) function: $f(X)=100\times \exp(x)+100\times \exp(2x)$,

where the expected value of $x$ is $0.05$ and standard deviation is $0.1$. normal distribution of x is assumed.

Community
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Toby
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  • The prime $'$ in $f'$ and $f'' = (f')'$ stands for the derivative. Wikipedia lists some basic formulas and rules for computing it, but if you're unfamiliar with this concept, it would be best to study some Calculus: the equivalent of a quarter of the first semester in a standard course would be enough. – whuber Jan 08 '13 at 16:17

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Can you not use $e^x \approx 1+ x+ \frac{X^2}{2} $ then $Var(e^X) \approx Var(X)+ \frac{Var(X^2)}{2} $ ?

It seems to me you want 2nd order approximation ..

dfhgfh
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  • how would you apply this on the above function? (normal distribution of x is assumed) – Toby Jan 08 '13 at 11:44
  • You just use the approximation I gave you, in other words you replace $e^x$ by $1+x+\frac{x^2}{2}$ and $e^2x$ by $1+2x+\frac{(2x)^2}{2}$. I coded this up and it works! – dfhgfh Jan 08 '13 at 12:49
  • and what would the formula for the variance look like? It is the 2x in exp(2x) and how this affects the Var(x^2) term in your formula that confuses me... – Toby Jan 08 '13 at 13:05
  • Bear in mind this is an approximation I am talking about. let $X$ be a $\mathcal{N}(0.05,0.1)$ then you can do the following : $100e^{X}+100e^{2X} \approx 100(1+X+\frac{X^2}{2} +1+2X+\frac{(2X)^2}{2} $ Now you apply $var$ in both sides and you get $var(100e^{X}+100e^{2X}) \approx 10000(9var(X) + \frac{25 var(X)}{4})$ – dfhgfh Jan 08 '13 at 15:17
  • @Toby If it could be relevant, why is the assumption of a normal distribution not mentioned in the question? – Glen_b Jan 08 '13 at 15:34
  • good point, I updated the question! – Toby Jan 09 '13 at 12:07