How can I plot the square root of normally distributed data?

I have a Gaussian distributed data with mean zero. I first want to take the square root of that data and then trying to find the standard deviation. I have the following issue with this:

By taking the square root of the data I am getting some imaginary numbers which are expected, so I am converting all data into real numbers by taking real(data)-imag(data). But if I look at the histogram of that data, it is no longer a Gaussian. I was expecting it to be Gaussian because if we take the square root of a Gaussian function it should be a Gaussian with increased standard deviation. I am not sure what am I missing here? A sample code is as follows

x= randn(1000); figure;histfit(x(1,:)); y=sqrt(x(1,:));
y1=real(y)-imag(y);figure;histfit(y1(1,:))

If the data is not normally distributed, is there any way to convert it into equivalent normal distribution? I am interested in three sigma rule of normal distribution.

I would be really thankful for suggestions.

Best regards

Ashok

[Walter Roberson]

I have never seen real(y)-imag(y) used to convert numbers to real. I would not expect the result to be gaussian.

Answer 1

[John Chilleri]

Hello,

The problem is that transforming data with a square root does not necessarily maintain the normal distribution, this is especially true if you consider numbers in the -1 to 1 range, as their square roots increase their magnitude. So a normal distribution around 0 would appear bimodal after taking the square root.

With large numbers, a simple, somewhat "solution" would be:

Data = sign(Data).*sqrt(abs(Data));

This maintains the signs of the Data, while square rooting their magnitudes.

The distribution will change because it was square rooted, but it will retain its signage, while doing what I believe you desire.

Hope this helps!

[AK]

Thanks a lot John.

But I am not sure why the rule of taking square root of a Gaussian function say exp(-x^2) is Gaussian doesn't apply to the normally distributed data. I would really appreciate your help.

[Walter Roberson]

exp(-x^2) can never be negative.

[AK]

Thanks Walter!

Do you have any suggestion about how can I report the error ( three sigma) in such distribution.