EECS 126 - Probability and Random Processes - J. Walrand

GAUSSIAN RANDOM VARIABLES

·         Key ideas

·         Gaussian

·         Jointly Gaussian

·         Conditional Expectation

 

Key Ideas

 

Gaussian random variables show up frequently.  (This is because of the central limit theorem that we discuss later in the class.) 

The Gaussian distribution is determined by its mean and variance.

The sum of independent Gaussian random variables is Gaussian. 

Random variables are jointly Gaussian if an arbitrary linear combination is Gaussian.

Uncorrelated jointly Gaussian random variables are independent.

If random variables are jointly Gaussian, then the conditional expectation is linear.

 

Gaussian

 

N(0,1): Standard Gaussian Random Variable

 

X = N(0, 1) if fX(x) = (1/2p)1/2exp{- x2/2}.  Equivalently, if E(exp{iuX}) = exp(- u2/2). 

 

Since E(exp{iuX}) = E(1 + iuX – u2X2/2 – iu3X3/3! + … + (iuX)n/n! + …)

                     = exp(- u2/2) = 1 - u2/2 + u4/8 – u6/48 … + (- u2/2) m/m! + …

   we see that E(X m) = 0 for m odd and

                    (iu)2mE(X 2m)/((2m)!) = (- u2/2) m/m!, so that E(X 2m) = {(2m)!}/{2mm!}

 

The c.d.f. of X does not admit a closed form expression.  It is useful to note that P(|X| > 2.6) = 1%, P(|X| > 2) = 5%, and P(|X| > 1.7) = 10%.

 

N(m, s2):

 

X = N(m, s2) if X = m + sY where Y = N(0,1), so that E(exp{iuX}) = exp(ium - u2 s2/2). 

 

The p.d.f. of a N(m, s2) is [1/(2p s2)]1/2 exp{- (x – m)2/2 s2).

 

Jointly Gaussian

 

N(0, I):

 

X = N(0, I) if the components of X are i.i.d. N(0,1). Then AX = N(0, AAT).

 

Jojntly Gaussian:

Random variables X are said to be jointly Gaussian if aTX is Gaussian for any vector a. 

If aTX  =  N, then E(exp{i aTX }) = exp{i aT m - aTSa}where S = cov(X,X), so that the components of X are independent if and only if is S diagonal, i.e., if and only if the random variables are uncorrelated.

Conditional Expectation

 

Let (X, Y) be jointly Gaussian.  Then we can look for a and B so that X – a – BY is zero-mean and uncorrelated with Y.  In that case, these random variables are independent and we can write

E[X | Y] = E[X – a – BY + a + BY | Y] = a + BY.

To find the desired a and B, we solve E[X – a – BY] = 0 and E[(X – a – BY)YT] = 0. This gives E[X|Y] = E(X) + M(Y - E(Y)) where M = cov(X,Y)cov(Y,Y)-1, assuming that cov(Y,Y) is invertible. If it is not, then we choose a pseudo-inverse cov(Y,Y)† such that

cov(X,Y) = cov(Y,Y)†cov(Y,Y).

 

Jean Walrand – March 2003  --- INDEX