# Expectation of deterministic signals

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!

## 3 Answers

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.

• I don't get the relation to the question. In the OP case there is no PDF to take the expectation with relate to. Hence it has no meaning if it is a Signal over time or constant. It just doesn't define the probability space. – Royi Jun 30 '18 at 23:52
• taking a rv to a limit is not particularly conceptually difficult or is using the delta function as a pdf. see Papoulis for examples. The answer was also prefaced as “hand waving”. Fundamentally the concept of the rv as passing to to a deterministic is not my invention, Ive seen an IEEE fellow, a student of Ziv make a similar intuitive augment . I don’t get what you don’t get. Not only that, it is a safe heuristic. Feel free to answer the OP and explain what a sigma algebra is to the OP. – user28715 Jul 1 '18 at 0:32
• I'm aware of creating a Delta Based Distribution (Some use that framework for Discrete Random Variables). I don't think it is related to the above case. for instance, let's say the noise depends on the deterministic signal so if you create a distribution for that the move you made with the Expectation doesn't work. It is simple, for the OP case there is no defined distribution for the Signal as it is not random hence the Expectation Operator has no effect related to it. – Royi Jul 1 '18 at 8:13
• I need a more concrete example of why “it doesn’t” work. since when does independence not work? – user28715 Jul 1 '18 at 12:24

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 $E[x(t)Y(t)]=x(t)E[Y(t)]$.

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.