skew normal distribution We have skew normal distribution with location $=0$, scale $=1$ and shape $=0$ then it is same as standard normal distribution with mean $0$ and variance $1$. But if we change the shape parameter say shape $=5$ then mean and variance also changes. How can we fix mean and variance with different values of shape parameter?
If we have $3$ equation of mean, variance and skewness then how can we fix location, scale and shape parameter. Can you explain about it?
 A: @Ross Millikan is right: you should look after and understand the background. But as I have answered this question on stackoverflow, let me answer it here also.
Just look after how the mean and variance of a skew normal distribution can be computed and you got the answer! Knowing that the mean looks like:
    and     
You can see, that with a xi=0 (location), omega=1 (scale) and alpha=0 (shape) you really get a standard normal distribution (with mean=0, standard deviation=1):

If you only change the alpha (shape) to 5, you can except the mean will differ a lot, and will be positive. If you want to hold the mean around zero with a higher alpha (shape), you will have to decrease other parameters, e.g.: the omega (scale). The most obvious solution could be to set it to zero instead of 1. See:

Mean is set, we have to get a variance equal to zero with a omega set to zero and shape set to 5. The formula is known:

With our known parameters:

Which is insane :) That cannot be done this way. You may also go back and alter the value of xi instead of omega to get a mean equal to zero. But that way you might first compute the only possible value of omega with the formula of variance given.

Then the omega should be around 1.605681 (negative or positive).
Getting back to mean:

So, with the following parameters you should get a distribution you was intended to:

location = 1.256269 (negative or positive), scale = 1.605681 (negative or positive - the opposite sign of location) and shape = 5.

A: If you follow the notation in the Wikipedia article you have the mean is $\xi + \omega \delta \sqrt{\frac{2}{\pi}}$ and the variance is $\omega^2(1-\frac{2\delta^2}{\pi})$ where $\delta=\frac{\alpha}{\sqrt{1+\alpha^2}}$.  So if you pick your $\alpha$ you can calculate $\delta$.  Your desired variance gives you $\omega$ and finally your desired mean gives $\xi$
