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I have never taken probability theory, and I wonder whether one can express some random variable $X$ with mean $\mu$ and variance $\sigma ^2$ in terms of $\mu$ and $\sigma$ only. Or at least something close to it.

Thank you for your help.

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    I'd first remark that upper case Roman letters (eg. $X$) are probably a less confusing notation for random variables. Also, I don't quite understand your question; do you want $\mu$ and $\sigma $ to be the sole parameters of $x$'s distribution?2012-06-13
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    You cannot express a _random variable_ in terms of its mean and variance, but you can express its _density function_ or distribution function etc in terms of the mean and variance. For example, the density function of a Gaussian (a.k.a. normal) random variable can be expressed in terms of its mean and variance: $f(x) = (2\pi\sigma^2)^{-1/2}\exp((x-\mu)^2/2\sigma^2)$.2012-06-13
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    @TimDuff if it is possible. Thats my question. Is it possible and if it is not then what is the closest way of doing it?2012-06-13
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    @DilipSarwate well. I saw somewhere that $$Z=\frac{X−μ}{σ}$$2012-06-13
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    Your $Z$ will have mean $0$ and standard deviation $1$, telling you its location and scale but not its shape. It will not have a normal distribution (or any other particular shape of distribution) unless $X$ does.2012-06-14

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No - the mean and variance (if they exist) tell you about the location and scale of a distribution but nothing about its shape.

If a random variable $X$ has mean $\mu$ and variance $\sigma^2$ then the random variable $Y$ defined by $Y=aX+b$ for real $a$ and $b$ has mean $a\mu+b$ and variance $a^2 \sigma^2$.

So given a particular shape of distribution (well behaved enough to have a mean and variance), it is possible to find a distribution which has the same shape but any mean and (positive) variance you specify.

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A Poisson random variable has mean lambda and variance lambda. In the case of one parameter distirbutions the mean and variance will both be functions of a single parameter.