If $h(n)$ is the impulse response of the discrete LTI system, $x(n)$ is the white noise process with variance $\sigma_x^2$ and $y(n)$ is the output, prove that

$$\mu_y = \mu_x \, \sum_{k=0}^\infty h(k)$$

where $\mu$ represents the mean.

My approach:

$$\mu_y= \mathbb{E}[y(n)] =\mathbb{E}[x(n) \star h(n)]$$ Where, '$ \star$' means convolution
Now how can I proceed?


The standard meaning of white noise includes an insistence (whether implicit or explicit) that the mean is $0$. Thus, what you want to prove is trivially true: since $$Y[n] = \sum_{k=-\infty}^\infty h[n-k]X[k] = h\star X \big\vert_{n},$$ the linearity of expectation (the notion that $E[aX]=aE[X]$ and that the expectation of a sum is the sum of the expectations) gives us that $$\mu_Y[n] = E\left[\sum_{k=-\infty}^\infty h[n-k]X[k]\right] = \sum_{k=-\infty}^\infty h[n-k]E\big[X[k]\big]$$ where all the $E\big[X[k]\big]$ have value $0$, and so both sides have value $0$.

More generally, if the $X$ process has constant mean $\mu_X$, then $$\mu_Y[n] = \sum_{k=-\infty}^\infty h[n-k]E\big[X[k]\big] = \sum_{k=-\infty}^\infty h[n-k]\mu_X = \mu_X\sum_{k=-\infty}^\infty h[n-k] = \mu_X \sum_{i=-\infty}^\infty h[i]$$ is also a constant.

  • $\begingroup$ How $$E[\sum {h(n-k) * X(k)}] = \sum {h(n-k) * E[X(k)]} $$ $$?$$ $\endgroup$
    – Suresh
    Feb 17 '18 at 11:45
  • 1
    $\begingroup$ Linearity of expectation means that $\displaystyle E\left[\sum_k a_kX_i\right] = \sum_k a_k E[X_k]$. It is a direct consequence of the generalized law of the unconscious statistician. $\endgroup$ Feb 18 '18 at 3:14

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