假设我们有随机变量,其方差和均值已知。问题是:对于给定的函数f ,的方差是多少。我知道的唯一通用方法是增量方法,但它仅提供近似值。现在我对f (x )= √感兴趣,但是了解一些通用方法也很高兴。
编辑29.12.2010
我已经使用泰勒级数进行了一些计算,但是我不确定它们是否正确,因此如果有人可以确认它们,我将非常高兴。
首先,我们需要近似
现在我们可以近似
使用的近似我们知道˚F (μ )- ë ˚F (X )≈ - 1
使用此,我们得到:
假设我们有随机变量,其方差和均值已知。问题是:对于给定的函数f ,的方差是多少。我知道的唯一通用方法是增量方法,但它仅提供近似值。现在我对f (x )= √感兴趣,但是了解一些通用方法也很高兴。
编辑29.12.2010
我已经使用泰勒级数进行了一些计算,但是我不确定它们是否正确,因此如果有人可以确认它们,我将非常高兴。
首先,我们需要近似
现在我们可以近似
使用的近似我们知道˚F (μ )- ë ˚F (X )≈ - 1
使用此,我们得到:
Answers:
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
So for any we can find large enough so that
First let us estimate . We have
Since the domain of the first integral is interval which is bounded closed interval we can apply Taylor expansion:
Substituting this formula to the previous one we get
Now for the variance we can use Taylor approximation for , subtract the formula for and square the difference. Then
where involves moments for . 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 :
Similarly we get then
The formula for variance then becomes
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.