Variance of a function of one random variable

Lets say we have random variable $X$ with known variance and mean. The question is: what is the variance of $f(X)$ for some given function f. The only general method that I'm aware of is the delta method, but it gives only aproximation. Now I'm interested in $f(x)=\sqrt{x}$, but it'd be also nice to know some general methods.

Edit 29.12.2010
I've done some calculations using Taylor series, but I'm not sure whether they are correct, so I'd be glad if someone could confirm them.

First we need to approximate $E[f(X)]$
$E[f(X)] \approx E[f(\mu)+f'(\mu)(X-\mu)+\frac{1}{2}\cdot f''(\mu)(X-\mu)^2]=f(\mu)+\frac{1}{2}\cdot f''(\mu)\cdot Var[X]$

Now we can approximate $D^2 [f(X)]$
$E[(f(X)-E[f(X)])^2] \approx E[(f(\mu)+f'(\mu)(X-\mu)+\frac{1}{2}\cdot f''(\mu)(X-\mu)^2 -E[f(X)])^2]$

Using the approximation of $E[f(X)]$ we know that $f(\mu)-Ef(x) \approx -\frac{1}{2}\cdot f''(\mu)\cdot Var[X]$

Using this we get:
$D^2[f(X)] \approx \frac{1}{4}\cdot f''(\mu)^2\cdot Var[X]^2-\frac{1}{2}\cdot f''(\mu)^2\cdot Var[X]^2 + f'(\mu)^2\cdot Var[X]+\frac{1}{4}f''(\mu)^2\cdot E[(X-\mu)^4] +\frac{1}{2}f'(\mu)f''(\mu)E[(X-\mu)^3]$
$D^2 [f(X)] \approx \frac{1}{4}\cdot f''(\mu)^2 \cdot [D^4 X-(D^2 X)^2]+f'(\mu)\cdot D^2 X +\frac{1}{2}f'(\mu)f''(\mu)D^3 X$

• Delta method is used for asymptotic distributions. You cannot use when you have only one random variable. Dec 28, 2010 at 14:35
• @mpiktas: Actually I dont know much about Delta method, I've just read something on wikipedia. This is quotation from wiki: "The delta method uses second-order Taylor expansions to approximate the variance of a function of one or more random variables". Dec 29, 2010 at 13:08
• it seems wikipedia has exactly what you want: en.wikipedia.org/wiki/…. I will reedit my answer, it seems that I underestimated Taylor expansion. Dec 29, 2010 at 13:39
• Tomek, if you disagree with the edits that were made (not by me), you can always change them again, or roll them back, or just point out the differences and ask for clarification. May 11, 2013 at 22:16
• @Glen_b: I agree with them E(X-mu) = 0 doesn't implyt that E[(X-mu)^3] = 0. May 12, 2013 at 7:32

Update

I've underestimated Taylor expansions. They actually work. I assumed that integral of the remainder term can be unbounded, but with a little work it can be shown that this is not the case.

The Taylor expansion works for functions in bounded closed interval. For random variables with finite variance Chebyshev inequality gives

$$P(|X-EX|>c)\le \frac{\operatorname{Var}(X)}{c}$$

So for any $$\varepsilon>0$$ we can find large enough $$c$$ so that

$$P(X\in [EX-c,EX+c])=P(|X-EX|\le c)<1-\varepsilon$$

First let us estimate $$Ef(X)$$. We have \begin{align} Ef(X)=\int_{|x-EX|\le c}f(x)dF(x)+\int_{|x-EX|>c}f(x)dF(x) \end{align} where $$F(x)$$ is the distribution function for $$X$$.

Since the domain of the first integral is interval $$[EX-c,EX+c]$$ which is bounded closed interval we can apply Taylor expansion: \begin{align} f(x)=f(EX)+f'(EX)(x-EX)+\frac{f''(EX)}{2}(x-EX)^2+\frac{f'''(\alpha)}{3!}(x-EX)^3 \end{align} where $$\alpha\in [EX-c,EX+c]$$, and the equality holds for all $$x\in[EX-c,EX+c]$$. I took only $$4$$ terms in the Taylor expansion, but in general we can take as many as we like, as long as function $$f$$ is smooth enough.

Substituting this formula to the previous one we get

\begin{align} Ef(X)&=\int_{|x-EX|\le c}f(EX)+f'(EX)(x-EX)+\frac{f''(EX)}{2}(x-EX)^2dF(x)\\\\ &+\int_{|x-EX|\le c}\frac{f'''(\alpha)}{3!}(x-EX)^3dF(x) +\int_{|x-EX|>c}f(x)dF(x) \end{align} Now we can increase the domain of the integration to get the following formula

\begin{align} Ef(X)&=f(EX)+\frac{f''(EX)}{2}E(X-EX)^2+R_3\\\\ \end{align} where \begin{align} R_3&=\frac{f'''(\alpha)}{3!}E(X-EX)^3+\\\\ &+\int_{|x-EX|>c}\left(f(EX)+f'(EX)(x-EX)+\frac{f''(EX)}{2}(x-EX)^2+f(X)\right)dF(x) \end{align} Now under some moment conditions we can show that the second term of this remainder term is as large as $$P(|X-EX|>c)$$ which is small. Unfortunately the first term remains and so the quality of the approximation depends on $$E(X-EX)^3$$ and the behaviour of third derivative of $$f$$ in bounded intervals. Such approximation should work best for random variables with $$E(X-EX)^3=0$$.

Now for the variance we can use Taylor approximation for $$f(x)$$, subtract the formula for $$Ef(x)$$ and square the difference. Then

$$E(f(x)-Ef(x))^2=(f'(EX))^2\operatorname{Var}(X)+T_3$$

where $$T_3$$ involves moments $$E(X-EX)^k$$ for $$k=4,5,6$$. We can arrive at this formula also by using only first-order Taylor expansion, i.e. using only the first and second derivatives. The error term would be similar.

Other way is to expand $$f^2(x)$$: \begin{align} f^2(x)&=f^2(EX)+2f(EX)f'(EX)(x-EX)\\\\ &+[(f'(EX))^2+f(EX)f''(EX)](X-EX)^2+\frac{(f^2(\beta))'''}{3!}(X-EX)^3 \end{align}

Similarly we get then \begin{align*} Ef^2(x)=f^2(EX)+[(f'(EX))^2+f(EX)f''(EX)]\operatorname{Var}(X)+\tilde{R}_3 \end{align*} where $$\tilde{R}_3$$ is similar to $$R_3$$.

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} where $$\tilde{T}_3$$ have only third moments and above.

• I dont need to know the exact value of the variance, approximation should works for me. Dec 29, 2010 at 13:10
• Indeed, the approximate formula for $\mathbb{E}[f(X)]$ in the OP is often used in risk analysis in economics, finance and insurance. Dec 29, 2010 at 14:50
• @Tomek Tarczynski, the third derivative of $\sqrt{x}$ goes to zero quite quickly for large $x$, but is unbounded near zero. So if you picked uniform distribution with support close to zero, the remainder term can get large. Dec 30, 2010 at 9:04
• Note that in your link the the equality is approximate. In this answer all the equations are exact. Furthermore for the variance note that the first derivative is estimated at the $EX$, not $x$. Also I never stated that this will not work for $\sqrt{x}$, only that for $\sqrt{x}$ the approximate formula might have huge error if $X$ domain is close to zero. Jun 28, 2017 at 8:55
• In your answer, you wrote \begin{align} f(x)=f(EX)+f'(EX)(x-EX)+\frac{f''(EX)}{2}(x-EX)^2+\frac{f'''(\alpha)}{3}(x-EX)^3 \end{align}. But isn't the denominator of the last term is $3!$? Jun 28, 2017 at 13:50

To know the first two moments of X (mean and variance) is not enough, if the function f(x) is arbitrary (non linear). Not only for computing the variance of the transformed variable Y, but also for its mean. To see this -and perhaps to attack your problem- you can assume that your transformation function has a Taylor expansion around the mean of X and work from there.