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Home/ Questions/Q 3676550
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T03:06:44+00:00 2026-05-19T03:06:44+00:00

we have skew normal distribution with location=0, scale =1 and shape =0 then it

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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

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  1. Editorial Team
    Editorial Team
    2026-05-19T03:06:45+00:00Added an answer on May 19, 2026 at 3:06 am

    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:

    alt text    and    alt text

    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):

    alt text

    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:
    alt text

    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:

    alt text

    With our known parameters:

    alt text

    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.

    alt text

    Then the omega should be around 1.605681 (negative or positive).

    Getting back to mean:

    alt text

    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) and shape = 5.

    Please, someone test it, as I might miscalculated somewhere with the given example.

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