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Calculate the pdf of the sum $Z[n] = 3/4^{(n-1)}X[1] + 3/4^{(n-2)}X[2] + ... + 3/4X[n-1] + X[n]$.

Where $X[n]$ is a $IID$ gaussian stochastic process with $mean=0$ and $variance =1$.

Thanks!

Farhad
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Junior
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1 Answers1

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For two random variables $x_1$ and $x_2$ having means $u_1$ and $u_1$, and variances $v_1$ and $v_1$

if $z= a \cdot x_1 + b \cdot x_2$

then, $\operatorname{mean}(z) = a\cdot u_1 + b\cdot u_2$ and $\operatorname{variance}(z) = a^2 \cdot v_1 + b^2 \cdot v_2$

Hence, for your problem, $\operatorname{mean}(Z) = 0$. $\operatorname{variance}(Z) = (3/4)^(2n-2) + (3/4)^(2n-4) + \cdots 1$, which is a simple geometric sum.

Now that you have the mean and variance of $Z$, the pdf will simply be a Gaussian Distribution with these calculate values of mean and variance.

user3675152
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