The question might be trivial, but I would like someone to correct me or confirm it.

I know that the Normal (Gaussian) distribution is completely determined by its mean and variance, but does that hold for any other distribution? I assume that the answer is no. I could notice this is true for many distributions, but there are some exceptions. For example, mean and variance are undefined for the Cauchy distribution.

  • $\begingroup$ What do you mean exactly with "described by their mean and variance" ? Are you asking if, for a given family of densities (say Normal distributions) $f_\theta$, the mapping $\theta \mapsto f_\theta$ is injective ? In my example $\theta \in \mathbb R\times \mathbb R_+$, $\theta = (\mu, \sigma^2)$. $\endgroup$ – Gabriel Romon Nov 3 '17 at 8:17
  • $\begingroup$ Let's take the Normal distribution as an example. I wanted to say that if you know the mean and variance of Normal distribution, then you know everything about it. In other words, the shape of the pdf is known. I wonder if this is the case with other distributions? I assume it is not, but I would like someone to confirm that. $\endgroup$ – Veljko Nov 3 '17 at 8:35
  • $\begingroup$ It depends on what you classify as description/nature of the distribution. In your example the distribution is known on base of its mean, its variance and the fact that it is normal (description). You could go for a description that includes everything we don't know yet if we only know mean and variance. $\endgroup$ – drhab Nov 3 '17 at 8:43
  • 3
    $\begingroup$ In general, there are infinitely many probability distributions that fit a given mean $m$ and variance $\sigma^2$. $\endgroup$ – Michael Nov 3 '17 at 8:55
  • 2
    $\begingroup$ If you know the form of distribution, i.e. normal, gamma, Poisson beforehand then you need to know the number of moments equal to the number of unknown parameters in the distribution. For instance Poisson only has one parameter so you just need to know the mean. For normal, as you say, you need mean and variance. Most common distributions only have one or two parameters but there are plenty with more. $\endgroup$ – user121049 Nov 3 '17 at 9:25

Sure, it holds for other distributions. Usual examples beyond the Normal, are the Exponential and the Poisson. The only parameter in both distributions is defined by the mean and variance (that are related in both cases). For the Chi-squared distribution, the existing parameter, called "degrees of freedom", is the the mean, and the variance is twice the value of this parameter.

But this property does not hold for all (or most) distributions, as you correctly stated. Some distributions don't even have finite expected value and/or variance defined. You can easily check, for instance, that a simple distribution such as $$f(x)=\frac{1}{x^2}\ \ \text{for}\ \ x\in[1,\infty), 0\ \ \text{otherwise},$$ does not have finite expected value, as the integral used to compute it does not converge.

In some cases, expressing the distribution with its parameters being the expected value and variance (or standard deviation) might require some thinking. For instance, you can express the Uniform distribution, more often expressed by just stating the limits $a$ and $b$, by $$f(x)=\frac{1}{2\sigma \sqrt{3}}\ \text{for}\ \ x\in [\mu-\sigma\sqrt{3},\mu+\sigma\sqrt{3}]\ \text{and}\ 0\ \text{otherwise}.$$ An interesting case is this: if $X$ has Log-Normal($\mu$,$\sigma$) distribution, the parameters $\mu$ and $\sigma$ which appears in the usual representation are not the expected value and standard deviation for $X$. They are the expected value and variance for $Y=\ln X$. In this case, however, you can express $\mu$ and $\sigma$ in terms of $E(X)$ and $V(X)$.


Check out any family of distributions which is parametrized by at least three parameters, e.g.

$$f_{a,b,c}(x)=N(a,b,c)\cdot (ax^2+bx+c)$$

for $x\in[0,1]$ and $N(a,b,c)$ some normalization factor. There are several combinations of parameters $a,b$ and $c$ which generate the same mean and variance. You can use higher moments to fix more than two parameters though.


Fix $m \in \mathbb{R}$ and $\sigma^2>0$. For any $p \in (0,1]$, define random variable $X$ by $$X = \left\{ \begin{array}{ll} m &\mbox{ with prob $1-p$} \\ m+ \frac{\sigma}{\sqrt{p}} & \mbox{ with prob $p/2$} \\ m-\frac{\sigma}{\sqrt{p}} & \mbox{ with prob $p/2$} \end{array} \right.$$ Then $X$ has mean $m$ and variance $\sigma^2$. The probability distribution for $X$ is different for each $p \in (0,1]$. Hence, there are an infinite number of probability distributions that give rise to mean $m$ and variance $\sigma^2$.

  • $\begingroup$ Any symmetric distribution with finite variance, when multiplied by a suitable constant, has mean 0 and variance 1. In view of this one doesn't expect distributions to be determined by mean and variance. $\endgroup$ – Kavi Rama Murthy Nov 3 '17 at 9:12
  • $\begingroup$ You mean "finite and positive variance" (excluding the variance-0 counterexample). Any random variable $Y$ with finite mean and variance $E[Y]$, $Var(Y)>0$ can be transformed into a mean-$m$, variance-$\sigma^2$ random variable $Z$ via $$Z=m+\frac{\sigma}{\sqrt{Var(Y)}}(Y-E[Y])$$ $\endgroup$ – Michael Nov 3 '17 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.