# Random Signals - statistical properties are time dependant?

I'm taking a course on DSP and we're being introduced to the random signals, in particular continuous time and discrete time random signals.

We're told that if we repeat a single random experiment at each point in time, the random variable associated with each random experiment when collected together will form the random signal. We are then told that the random signal/process has statistical properties such as mean that are time-dependant.

Now that makes no sense. Surely, if you're doing the same experiment, the mean of the entire random signal will be the same as the mean of the random variable for one experiment? How will time affect the random signal?

• Could you elaborate a little bit. I would also agree that any time dependent property of a random signal would also be time dependent, but I don’t see what the distinction is between the random variable and random signal in this case. – Dan Szabo Oct 24 '18 at 13:39

## The answer to the question (a counterexample)

Properties of random processes will in general be time-dependent. They are not only when talking about stationary processes. Another related concept (not relevant here but that you might find interesting) is ergodicity. Most of us find it difficult to understand the difference between both concepts, but then you find this brilliant Dilip's answer to clear out all doubts.

Going back to your question: take for example the simple random process

$$X(t)=At$$

where $$A$$ is some random variable with mean $$\mu_a$$. The mean of the process would then be:

$$\mathbb{E}[X(t)]=\mathbb{E}[At]=t\mathbb{E}[A]=t\mu_a$$

It's easy to see that the mean depends on $$t$$, thus it's time-dependent. I hope this easy example shows why, in general, properties of random processes vary with time.

## Additional info that might help

I think your confusion arises from thinking that to build a stochastic process you grab many realizations of a single random variable and put them in a sequence to represent a time-signal. That's not how a random process is built in general. If you do this, then the mean will not vary with time because the process will be stationary. But if you don't, it may or may not vary.

I'll use an example here to make myself clear. Suppose you have a random variable $$Z$$ that represents the result of throwing a fair 6-faced dice. Then $$\mu_z = 3.5$$. If you throw the dice 3 times and you put them together to form a discrete-time signal of 3 samples, then you will have the following random process:

$$X(n)=Z_1\delta(n) + Z_2\delta(n-1) + Z_3\delta(n-2)$$

where $$Z_i$$ represents the random variable corresponding to the $$i$$-th throw of the dice.

Note that each sample of $$X(n)$$ is a random variable. This is correct for any random process in general. For a given random process $$X(n)$$, $$X(n_0)$$ for any $$n_0$$ will be a random variable.

On the other hand, once you've thrown the dice 3 times, you are not dealing with anything random anymore: you have a signal. That is, a realization of the random process.

If you build a random process such as we have just built $$X(n)$$, then you are right: the mean of the process will be constant and equal to the mean of the random variable used for each sample (as in this case). In fact, all properties will be constant and equal to those of the variable. But that's just in that specific case, because all samples of the process will be i.i.d., and thus, as @Fat32 has explained in the comments, the random process will be independent of time ($$t$$). In our dice example:

\begin{align} \mathbb{E}[X(n)]&=\mathbb{E}[Z_1\delta(n) + Z_2\delta(n-1) + Z_3\delta(n-2)]\\ &=\mathbb{E}[Z_1]\delta(n) + \mathbb{E}[Z_2]\delta(n-1) + \mathbb{E}[Z_3]\delta(n-2)] \\ & = \mu_z \qquad \text{for any} \ n\in\{0,1,2\} \end{align}

But this will not always be the case with random processes, such as in the example I wrote at the top of the answer, in which every random variable obtained by evaluating the random process at different values of $$t$$ is different from all others, therefore preventing the process from being i.i.d. and making its statistical properties time-dependent.

• Blush! ${}{}{}$ – Dilip Sarwate Oct 24 '18 at 14:00
• @DilipSarwate I have read that answer of yours more times than I can even remember. It was impossible not to bring it up here! – Tendero Oct 24 '18 at 14:10
• (+1) Whenever the indexed random variables $X_k$ that make up a random process are the same, having exactly the same PDFs, then the associated random process will be independent of time (index). In the example you have given; $X(t) = A t$ the associated indexed random variables $X_t$ are different (they have different PDFs) for each $t$. So even though the same experiment is performed for each $t$, the mapping from sample space S into real line R (which defines the RV) is different and thus creates a different RV for each t. This point should better be emphasized. @DilipSarwate – Fat32 Oct 25 '18 at 13:52
• @Fat32 Thanks for your insight. I've added some lines to emphasize what you explained above. Let me know if it's still unclear and I'll try to make myself more clear. – Tendero Oct 25 '18 at 13:59
• No that's very nice and clear answer. The point, however, is if you read the OP's question (the 3rd paragraph), it seems that his main confusion stems from assuming that the same experiment would generate the same random variable for each index k. Which is not true as I have underlined. Hence your answer could focus on this at the beginning. Then you dont even have to mention about Dilip's answer, as there is no need. Once it's accepted that the same experiment can generate different RVs at each index, then the non-stationarity (time-dependence) will be a direct consequence without examples. – Fat32 Oct 25 '18 at 14:08