## mean of a function of a random variable

Hi,

I have a random variable X with some zero-mean distribution.

I have a function Y of this r.v. given by something complicated
$Y=(a+X)^\frac{2}{3}$

Is there an explicit way of finding the distribution of Y or even its mean?

Thanks

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 Quote by Apteronotus Hi, I have a random variable X with some zero-mean distribution. I have a function Y of this r.v. given by something complicated $Y=(a+X)^\frac{2}{3}$ Is there an explicit way of finding the distribution of Y or even its mean? Thanks
Hey Apteronotus and welcome to the forums.

The general expression to find a mean is given by E[g(X)] = integral g(x)f(x)dx where f(x) is the PDF and you integrate over the domain of the random variable. For discrete replace an integral with a summation.

In your case the g(x) = (a+x)^(2/3). So if you know f(x) and its continuous, then plug g(x) in and solve the integral. If its discrete then but g(x) and find the summation.

 In terms of finding the distribution of Y there are a few techniques. One technique is through transformation methods of the PDF which has to do with finding the distribution of Y = f(X) (i.e. find PDF of Y given Y = f(X)). Other methods that are good for really complicated expressions involve finding the moment generating function and then using the characteristic equation in probability theory to get the final PDF (for continuous variables). For your purpose, I would first look at the transformation theorems for PDF. Take a look at the following on page 4: http://www.math.ntu.edu.tw/~hchen/te...s/lecture7.pdf

## mean of a function of a random variable

Hi Apteronotus,

Besides the general methods explained by chiro to calculate the expected value and distributions, since in this case you have a zero mean distribution you can use Taylor to get a summatory expression of $Y$ and, since $E(X) = 0$, you have that $E(Y)=a^\frac{2}{3}$

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 Quote by viraltux since in this case you have a zero mean distribution you can use Taylor to get a summatory expression of $Y$ and, since $E(X) = 0$, you have that $E(Y)=a^\frac{2}{3}$
Well, yeah, more generally I am assuming $X,Y \in ℝ$, which the OP didn't say, but I think it is a fair assumption for this question.