Expected value and variance of the square root of a random variable

Let $X$ be a univariate continuous random variable for which I can calculate all raw and central moments.

Is there an exact way to calculate $E[\,\sqrt{X}\,]$ and $\mathrm{Var}[\,\sqrt{X}\,]$ in this case?

That is to say, I am interested in calculating or estimating $E[\,g(X)\,]$ and $\mathrm{Var}[\,g(X)\,]$ when $g$ is the square root function.

I have asked this in a general way here: Approximating the expected value and variance of the function of a (continuous univariate) random variable . I have also read answers and coments to this question: Variance of powers of a random variable , but I think it refers to integer powers, which is not my case.

• How are you obtaining all the moments? For example, are they coming from an MGF? Sep 7, 2017 at 12:04
• @Glen_b I have an (exact but long) expression for calculating the $n$-th raw and central moments of $X$. I don't think this is the same as the MGF, is it? Sep 8, 2017 at 7:25
• No, though it might possibly allow us to write one. How was the expression obtained? Sep 8, 2017 at 7:43
• @Glen_b I do not know how the expressions for the moments were obtained, but they are exact expressions that involve the Gamma function and things like that. I do know the exact expression for the PDF of $X$. Sep 8, 2017 at 7:45
• If you have the pdf we can try to use the law of the unconscious statistician. Sep 8, 2017 at 7:55

If you only have the sequence of moments, the $\frac12$-th moment is not necessarily determined by them.

If the MGF exists in a neighborhood of zero, then the moment sequence would determined the distribution and the $\frac12$-th moment should be determined (though not always amenable to algebraic calculation).

However if you have the pdf, we can avoid all that, since we can try to compute $E(X^\frac12)$ directly (e.g. by calculating the integral $\int_0^\infty x^\frac12 f(x) dx$ -- I presume $X$ is on the non-negative half line,for the obvious reason). Note also that the distribution of $\sqrt{X}$ will be very easy to write down (if $Y=\sqrt{X},\, F_Y(y)=F_X(y^2)$ and $f_Y(y)=2yf_X(y^2)$). If, for example, we recognize that density as a standard one - it might be very fast to identify the expectation that way.

As whuber points out in comments, all we need to find is $E(\sqrt{X})$, since $\text{Var}(\sqrt{X})=E(X)-E(\sqrt{X})^2$.

• +1. Most people would understand the "negative half line" to be $(-\infty, 0)$, though. For completeness you might consider pointing out that since $E[X]$ is included among the known moments, $E[\sqrt{X}]$ suffices for computing $\operatorname{Var}(\sqrt{X})=E[X]-\left(E[\sqrt{X}]\right)^2$.
– whuber
Sep 8, 2017 at 13:03
• @whuber Thanks. Sorry if I am being dense but I'm not sure I get the point you're making with first. I talk about the nonnegative halfline (to refer to $[0,\infty)$, since $\sqrt{X}$ and moments of $\sqrt{X}$ would be real there). The only issue I can see is that the third $n$-keypress in "nonnegative" didn't come out so it presently says "nonegative" though I presume that's not especially confusing. Your second point is well taken, though. Sep 9, 2017 at 0:19
• Sorry; I simply misread the "nonegative" (multiple times): my eyes must have converted it into "negative" rather than the intended "non-negative".
– whuber
Sep 9, 2017 at 13:25
• Thank you all for your contributions! @Glen_b, do you mean that what I am looking for is the $\frac{1}{2}$-th moment of $X$? Sorry for my poor knowledge about moments. Sep 18, 2017 at 16:04
• Yes, $E[X^\frac12]$ could reasonably be called the "$\frac12$-th moment", but it's not a commonly used term. I wanted a way to refer to it without writing algebra that would interrupt the flow of language both when writing and reading. Sep 18, 2017 at 23:22

if you are only interested in the upper bound of the expectation, you can use Jensen's Inequality to immediately upper bound $$E[\sqrt{X}] \leq \sqrt{E[X]}$$, if $$E[X]$$ is sufficiently close to 1, the approximation would be quite good..

Otherwise the standard tool to approximate the expectation is to use the Taylor series of $$y = X-1$$, $$\sqrt{1+y} = 1 - y/2 + y^2/8 - y^3/16 + 5y^4/128 \dots$$

and since you already have the the MGF of $$X$$, you should be able to calculate that quickly...

• Because the Taylor series has a radius of convergence of just $1,$ it is unreasonable to expect this to work unless the chance of $X$ lying beyond the interval $(0,2)$ is negligible.
– whuber
May 25, 2019 at 19:50