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I would like to find upper bound of variance of random variable X if I know that mean of X is equal to one. Could you help me, please? Thanks!

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    Unfortunately nothing can be said without more information. For example, you can have [normally distributed random variables](https://en.wikipedia.org/wiki/Normal_distribution) with mean 1 and any $\sigma$ – 2017-02-14

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Given that you only know the expectation value of $X$ you really can't tell much about the variance. You could for example have a variable that is $1$ (almost) certainly which has zero variance, and you could have a variable with normal distribution with any positive standard deviation.

So the only thing you now from that information is that the variance is either zero or greater than zero.

As you're looking for an upper bound there's even worse news. Consider for example a variable with the distribution:

$$\phi(x) = \begin{cases} x^{-3}/2 & \text{if } x \ge 1/2\\ 0 & \text{otherwise} \end{cases}$$

You have that $\int \phi(x)dx \int_{1/2}^\infty x^{-3}dx/2= 1$ and $E[X] = \int x\phi(x)dx = \int_{1/2}^\infty x^{-2} dx/2 = 1$. However if you try to calculate the variance you end up with $V[X] = \int (x^2-1)\phi(x)dx = \int_{1/2}^\infty (x^{-1} - x^{-3}) dx/2$ which doesn't converge. That is it's possible that the variance isn't even finite.

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    I thought it, but I hoped that I was wrong. Thank you. – 2017-02-14
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    I'll just add that in general, all central moments are independent. You can have known mean, kurtosis, and skewness, and still generate random variables with any conceivable variance. – 2017-02-14