6

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.

Vicent
  • 639
  • 1
  • 5
  • 18
  • 1
    How are you obtaining all the moments? For example, are they coming from an MGF? – Glen_b Sep 07 '17 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? – Vicent Sep 08 '17 at 07:25
  • No, though it might possibly allow us to write one. How was the expression obtained? – Glen_b Sep 08 '17 at 07:43
  • 1
    @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 $X$. – Vicent Sep 08 '17 at 07:45
  • The pdf is probably the best place to start, since if the moments increase in size too fast the MGF doesn't exist in a neighborhood of zero and then the half-th moment would not be uniquely determined by the integer moments.. On the other hand, sometimes the expression for the $k$th moment doesn't rely on $k$ being integer, so it's probably worth giving the expression if it's not too hard to write down. – Glen_b Sep 08 '17 at 07:50
  • 1
    If you have the pdf we can try to use the law of the unconscious statistician. – Glen_b Sep 08 '17 at 07:55

2 Answers2

7

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 $\text{Var}(\sqrt{X})=E(X)-E(\sqrt{X})^2$.

Glen_b
  • 257,508
  • 32
  • 553
  • 939
  • +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 08 '17 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 *non*negative 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. – Glen_b Sep 09 '17 at 00:19
  • Sorry; I simply misread the "nonegative" (multiple times): my eyes must have converted it into "negative" rather than the intended "non-negative". – whuber Sep 09 '17 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. – Vicent Sep 18 '17 at 16:04
  • 1
    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. – Glen_b Sep 18 '17 at 23:22
  • @Glen_b, Thank you for the answer. One moment that I miss, what is the distribution $f_Y(y)$ if $f_X$ is standard Normal in your example? – garej May 10 '19 at 08:06
  • To get a real-valued square root, $X$ must be non-negative; what situation would this (taking the square root of a standard normal) make sense for? – Glen_b May 10 '19 at 11:01
1

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...

Nick Lim
  • 41
  • 2
  • 1
    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 '19 at 19:50