Variance of function of random variable

Question

Is their an easier way to find variance of function of random variable?

Till now what I am doing is first find probability density function of (function of random variable) then integrate over range.

Answer 1

$$D(f(\xi)) = E((f(\xi) - Ef(\xi))^2) = Ef^2(\xi) - (Ef(\xi))^2$$ But often $f(\xi)$ has known distribution with known variance

Answer 2

Since $V(X)=E(X^2)-(E(X))^2$, and since for $Y=g(X)$ you have $$E(Y)=E(g(X))=\sum g(x) p_X(x)$$ (if $X$ is discrete, with $x$ taking all values for which $X$ has positive probability) or $$E(Y)=E(g(X))=\int_{-\infty}^\infty g(x) f_X(x) dx$$ (if $f$ is continuous with density $f_X$), you can find $V(Y)$ by finding $E(Y^2)=E((g(X))^2)$ and $E(Y)=E(g(X))$.

For instance, if $X$ is discrete, you would have $$E(Y^2)=E(g(X)^2)=\sum (g(x))^2 p_X(x)$$ and $$E(Y)=E(g(X))=\sum g(x) p_X(x).$$

Answer 3

(For general distributions): $Ef(X)=\int f(x)dF_X(x)$ and $E(f(X))^{2}=\int f(x)^{2}dF_X(x)$. So $var(f(X))=\int f(x)^{2}dF_X(x)-(\int f(x)dF_X(x))^{2}$.