Variance of a summation of Gaussians

  • Thread starter Thread starter SeriousNoob
  • Start date Start date
  • Tags Tags
    Summation Variance
SeriousNoob
Messages
12
Reaction score
0

Homework Statement


I am trying to follow a step in the textbook but I don't understand.

var\left(\frac{1}{N}\sum_{n=0}^{N-1}w[n]\right)\\<br /> =\frac{1}{N^2}\sum_{n=0}^{N-1}var(w[n])
where w[n] is a Gaussian random variable with mean = 0 and variance = 1

Homework Equations



Var(X) = \operatorname{E}\left[X^2 \right] - (\operatorname{E}[X])^2.<br />

The Attempt at a Solution


The mean is 0 because a summation of Gaussian is Gaussian.
But squaring the whole expression doesn't seem right as there seems to be a trick used to go from line 1 to 2.
 
Last edited:
Physics news on Phys.org
Two facts are being used here:

1. If X is a random variable and c is a constant, then \text{var}(cX) = c^2\text{var}(X).
2. If X and Y are uncorrelated random variables, then \text{var}(X + Y) = \text{var}(X) + \text{var}(Y). From this, it's an easy induction to handle the sum of N uncorrelated random variables.

Both of these facts are straightforward to prove and should be found in any probability book.
 
Thanks a lot. Haven't touched random variables for a while and the summation threw me off.

The proofs for those facts are indeed very straightforward.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top