How to fit distribution having expected value and 0.95 quantile? I would like to estimate parameters of lognormal distribution having only two values: expected value and 0.95 quantile. What do you recommend? Thanks!
 A: Let the parameters be $\mu$ and $\sigma$, which are the expectation and standard deviation of the associated normal distribution.  Then the expected value of the lognormal distribution is $m = \exp(\mu + \sigma^2/2)$ and its $0.95$ quantile is $q = \exp(\mu + z_{0.95} \sigma)$ where $z_{0.95}$ is the $0.95$ quantile of the standard Normal distribution (equal approximately to $1.645$).  Taking logarithms yields two simultaneous equations
$$\log(m) = \mu + \sigma^2/2$$
$$\log(q) = \mu + z_{0.95} \sigma.$$
There are two solutions,
$$\sigma = z\pm\sqrt{z^2+2 \log\left(\frac{m}{q}\right)};\quad \mu = \log(q) - z\sigma.$$
($z_{0.95}$ is abbreviated $z$.)
Only one of them gives a positive value of $\sigma$ unless $q \gt m$.  (Lognormal distributions can be so skewed that their means exceed high upper percentiles.)  For example, with $m=1$ and $q=2$, the two PDFs have graphs like these:

You will need to provide some additional criterion for selecting between these two solutions.
A: First of all, if you have the theoretical values of the mean and the 0.95 quantile, then I would not say to estimate but to calculate. 
You have that the mean of a lognormal distribution is given by 
$$E[X]=\exp(\mu +\sigma^2/2),$$
and that the quantile $p$ is given by
$$F^{-1}(p)=\exp(\mu+\Phi^{-1}(0.95)\sigma),$$
where $\Phi^{-1}(0.95)\approx 1.64$. By using the values you have for the mean and the 0.95 quantile, say $(m,q)$, you can calculate $(\mu,\sigma)$ by solving the equations system $E[X]=m$ and $F^{-1}(p)=q$.
