I have a question related with statistic. I read some other website and they said that: "An easy way to approximate normal distribution is to add three random numbers:
G = X + X + X
X = a uniformly distributed random number between -1 and 1.
G ~ a standard normal random number."
and
"
If X and Y are independent Gaussian random variables with mean 0 and standard deviation sigma, then sqrt(X^2 + Y^2) has a Rayleigh distribution with parameter sigma."
Adding several iid (independent, identically distributed) RV (random variables) get a distribution that resembles more and more a normal distribution.
It is a consequence of the central limit theorem.
The number of RV to be added depends on how much similar to gaussian has to be the resultant distribution.
Only 3 RV give a rough approximation, but enough for many applications. It is usual to use 6. Using 10 or more is much better, in the sense that extreme values (values with very low probability) can be obtained only with many additions.
If X and Y are independent Gaussian random variables with mean 0 and standard deviation sigma, then sqrt(X^2 + Y^2) has a Rayleigh distribution with parameter sigma.
I know that we can get the gaussian approximation by central theory. However, in my question, why is three not four? I tried by matlab the four normal RV case (adding four normal RV) and using the histogram to plot the pdf of the sum. the result showed that not not the shape of gaussian RV. but in three normal RV case, the pdf shape is gaussian. I dont know why,
I tried by matlab the four normal RV case (adding four normal RV) and using the histogram to plot the pdf of the sum. the result showed that not not the shape of gaussian RV. but in three normal RV case, the pdf shape is gaussian. I dont know why,
How many samples have you used for the histogram? Remember that a sufficient number of samples is required in order that the histogram resembles the distribution. Please try to increase the number of samples.
Regards
I mean if I add 4 normal RV between -1 and 1, I cannot get a standard normal distribution ( mean of 0 and variance of 1).
>> a=rand(1,100000)*2-1;
>> b=rand(1,100000)*2-1;
>> c=rand(1,100000)*2-1;
>> d=rand(1,100000)*2-1;
>> x=a+b+c;
>> hist(x,100)
>> y=a+b+c+d;
>> figure
>> hist(y,100)
>> mean(x)
A RV with uniform distribution between -1 and 1 has mean µ=0 and variance σ²=1/3.
If n such RV are added, the resultant mean is nµ and variance nσ².
So, the results you got are consistent. In order to obtain variance 1, is is necessary to multiply the result by √(3/n) .
Regards