问题描述
我们有位置=0，尺度=1和形状=0的偏态正态分布，那么它与均值为0，方差为1的标准正态分布相同.但是如果我们改变形状参数，比如shape=5，那么均值和方差也发生了变化.我们如何修复具有不同形状参数值的均值和方差
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
推荐答案
只要看看如何计算偏态正态分布的均值和方差，您就会得到答案！知道平均值看起来像:
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:
和
and
您可以看到，当 xi=0(位置)、omega=1(比例)和 alpha=0(形状)时，您确实得到了标准正态分布(均值=0，标准差=1):
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):
如果您只将 alpha(形状)更改为 5，则除了平均值会有很大差异外，您还可以为正数.如果您想以更高的 alpha(形状)将均值保持在零附近，则必须减少其他参数，例如:omega(比例).最明显的解决方案可能是将其设置为零而不是 1.请参阅:
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:
均值已设置，我们必须得到方差为零，欧米茄设置为零，形状设置为 5.公式已知:
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:
使用我们已知的参数:
这太疯狂了 :) 不能这样做.您也可以返回并更改 xi 而不是 omega 的值以获得等于零的均值.但是那样你可能首先用给定的方差公式计算 omega 的唯一可能值.
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.
那么欧米茄应该在 1.605681 左右(负或正).
Then the omega should be around 1.605681 (negative or positive).
回到原意:
因此，使用以下参数，您应该得到您想要的分布:
So, with the following parameters you should get a distribution you was intended to:
位置 = 1.256269(负或正)，比例 = 1.605681(负或正)，形状 = 5.
请有人测试一下，因为我可能会用给定的例子在某处计算错误.
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