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Unformatted text preview: m variables is
again Gaussian, and the variance of the sum, σ 2 , is equal to the sum of the variances. The result
can be extended to the case that X1 and X2 have nonzero means µ1 and µ2 . In that case, the sum
is again Gaussian, with mean µ = µ1 + µ2 and variance σ 2 = σ1 + σ2 .
The above derivation is somewhat tedious. There are two more elegant but more advanced
ways to show that the convolution of two Gaussian pdfs is again a Gaussian pdf. (1) One way
is to use Fourier transforms–convolution of signals is equivalent to multiplication of their Fourier
transforms in the Fourier domain. The Fourier transform of the N (0, σ 2 ) pdf is exp(−σ 2 (2πf )2 /2),
so the result is equivalent to the fact:
exp(−σ1 (2πf )2 /2) exp(−σ2 (2πf )2 /2) = exp(−(σ1 + σ2 )(2πf )2 /2). (2) The other way is to use the fact that the the sum of two independent binomial random variables
with the same p parameter is again a binomial random variable, and then appeal to the DeMoivreLaplace limit theorem, stating Gaussian dist...
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- Spring '08
- The Land