Let $X$ be a positive, lognormal random variable with known mean $\mu_X$ and variance $\sigma_X^2$. Since $X$ is a lognormal random variable, I know its pdf and moment-generating function (mgf).

pdf: $$f(x) = \frac{1}{x\sigma\sqrt{2\pi}}\exp\left[-\frac{\ln[x-\mu_X]}{2\sigma^2}\right]$$

mgf: $$E[X^n] = \exp\left[n\mu+\frac{1}{2}n^2\sigma^2\right]$$

Now, what I'm interested in is not only mean and variance of $X$, but also of its square root $\sqrt{X} = X^{1/2}$.

I have read the posts Expected value and variance of the square root of a random variable and Variance of powers of a random variable and their comments (among others), where I learned that I could use the mgf to estimate $E[X^\frac{1}{2}]$. However, this required that I use the "$\frac{1}{2}$"-th moment. More specifically, compute derivative of the mgf at $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$, which I haven't been able to do, due to the "partial" nature of the derivative.

Similarly, I figured that I could use the pdf, by solving the integral.

$$\int_0^\inf \sqrt{x} f(x) dx$$

Again, I haven't been able to do so, and solutions derived using the typical toolg (e.g. https://www.integral-calculator.com/) haven't been helpful so far.

Do you guys have any pointers in how else I could tackle this (using attempts I described here, or completely novel ones).

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    $\begingroup$ You can use Mathematica for calculation. mathematica Clear["Global`*"]; (*Define the distribution*) distribution = LogNormalDistribution[\[Mu], \[Sigma]]; (*Compute the mean of the square root of X*) meanSqrtX = Mean[TransformedDistribution[Sqrt[x], x \[Distributed] distribution]] Mean[TransformedDistribution[x^n, x \[Distributed] distribution]] $$ X \sim \text{LogNormalDistribution}[\mu ,\sigma ] \\ \begin{align} \mathbb{E}[\sqrt{X}]& = e^{\frac{\mu }{2}+\frac{\sigma ^2}{8}} \\ \mathbb{E}[X^n] &= e^{\frac{\sigma ^2 | n| ^2}{2}+\mu n}\end{align} $$ $\endgroup$
    – 138 Aspen
    Commented Mar 5 at 6:59

2 Answers 2


The square root of a lognormal is itself lognormal. Indeed $Y^p$ is lognormal more generally.

Note that if $X\sim N(\mu,\sigma^2)$ then $Y=\exp(X)$ is lognormal with the same parameters and vice versa. I assume you're happy with obtaining the mean and variance of $Y$ from the parameters. Failing that they're specified in the wikipedia article on the lognormal (e.g. see the sidebar on the right if you're on a computer rather than a mobile phone).

Now $Y^{\frac12} = \exp(X/2)$ by basic properties of powers -- $(b^m)^n = b^{mn}$. But $X/2\sim N(\mu/2,\sigma^2/4)$. Hence $Y^{\frac12}$ is lognormal with parameters $\mu/2$ and $\sigma^2/4$. Its mean and variance follow immediately.

Essentially the same argument works for other powers of $Y$.

However, if you're estimating these parameters from a sample, beware the (downward) bias in estimation of the distribution's mean and variance from the MLEs of the parameters may be substantial, even when the samples are fairly large.

I'd also add a note of caution for later readers that this discussion applies for the ordinary two-parameter lognormal. Occasionally people say "lognormal" when referring to a shifted lognormal, and in that case the analysis in my discussion above would not work.

  • $\begingroup$ Oh that's great to know! I did come to the conclusion that their distribution has the same shape (if that's the correct term), but didn't make the connection that man and variance could be derived so easily - thanks! $\endgroup$ Commented Feb 27 at 14:43
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    $\begingroup$ Well, not the same shape per se, but I know what you're getting at. If you change $\sigma^2$, you change skewness (the shape definitely changes). The square root is less skew than what you started with; indeed, generally much less, unless the $\sigma^2$ parameter was already quite small. It is the same distribution family but that family includes anything from as near to symmetric as you like to extremely heavily skewed. $\endgroup$
    – Glen_b
    Commented Feb 27 at 15:39
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    $\begingroup$ Yes, I just realised that the same "shape" I was referring to was the shape of $ln(Y)$ and $ln(Y^2)$. That's clearly something else than those of $Y$ and $Y^2$ themselves - very misleading, my bad. $\endgroup$ Commented Feb 27 at 16:33

If Wikipedia is trustworthy enough for you, it says that

$$ E(X^n) = \text{exp}\big(n\mu+\frac{1}{2}n^2\sigma^2\big) $$

for real or complex $n$, so you can use this formula with $n=\frac{1}{2}$.

Simulations are consistent with this formula. R code, adapt as desired:

logmean <- 2
logsd <- 2
sims <- rlnorm(1e5,logmean,logsd)
  • $\begingroup$ This seems to do the trick, thank you! I'm not sure how, but this completely eluded me - but I am quite new to this kind of thinking, involving mfgs. $\endgroup$ Commented Feb 27 at 14:41
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    $\begingroup$ You don't have to trust Wikipedia: this formula is proven here at stats.stackexchange.com/questions/116644. Additional information can be found by searching our site. $\endgroup$
    – whuber
    Commented Feb 27 at 21:27

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