Expectation of deterministic signals

Question

Maybe this a stupid question, but I came along with this point and I would like to confirm that I am not wrong.

I found some similar posts in StackExchange, like:

Expectation and Auto-correlation of an Output from a Matched Filter.

In one of the answer, it is stated "as S(t) is determistic[...]" and then the deterministic signal is taken out of the expectation.

and we know that for instance if I have a constant value, I can take it out of the Expectation operator, e.g.:

Where lambda is a random variable and "a" is a constant. Ok, so, if "a" is a constant, then fair enough, but what happens when "a" is a deterministic signal, but it is not a constant, for instance, what happens if "a" is a squared subcarrier?

Can I still safely say that the expectation will be the product of the deterministic signal multiplied by the expectation of the random variable?

Thanks, guys!

Answer 1

A hand-waving example might be helpful, let's assume $x$ is Gaussian with mean $m$ variance $\sigma$

$$ E\{ x\}=m = \frac{1}{\sqrt{\pi}\sigma}\int_{-\infty}^{\infty} x \exp(-\frac{1}{2} \left( \frac{(x-m)}{\sigma} \right)^2 ) dx $$ $m$ doesn’t depend on $\sigma$ as long as $\sigma>0$. If we left $\sigma \Rightarrow 0$ $$ \frac{1}{\sqrt{\pi}\sigma} \exp(-\frac{1}{2} \left( \frac{(x-m)}{\sigma} \right)^2 )\; \Rightarrow \; \delta(x-m) $$ where $\delta(t)$ is the Dirac Delta, You can think of a deterministic variable as a special case of an infinite SNR random variable or perhaps that a deterministic variable has a pdf like: $$ p_x=\delta (x-m(t)) $$ So going back to your factoring out your constant, You should be ble to show without too much trouble, If $x$ and $y$ are independent, $$ E\{xy\}= E\{x\}E\{y\} $$ then $$ E\{a \lambda\} = E\{a\}E\{\lambda\}= aE\{\lambda\} $$ The trickiest part of Expectations is knowing when an expectation is a deterministic variable or a random variable. The conditional expectation, $$ E\{x | Y\} \; \text{where}\; Y \; \text{is a random variable, the expectation is a random variable} $$ $$ E_{y}\{E_{x} \{x | Y\}\} \;\text{ is a deterministic quantity.} $$ Unfortunately most books don't carry the subscripts of which quantity the expectation is operating on.

Answer 2

If $x(t)$ is a deterministic signal, then you have $E[x(t)]=x(t)$. Furthermore, if $Y(t)$ is a random process you have $E[x(t)Y(t)]=x(t)E[Y(t)]$.

Answer 3

When you see a PDF it's mostly asymptotic, If it's highly random, then it's more spread, and variance being higher. So intuitively a deterministic process would have a pdf as an Impulse function with amplitude tending to infinite and unit area.