# Why doesn't law of large numbers apply to this stationary time-series?

There's a paragraph in Wikipedia that states the following:

Let Y be any scalar random variable, and define a time-series $\{X_t\}$, by $$X_{t}=Y\qquad {\text{ for all }}t$$ Then $\{X_t\}$ is a stationary time series, for which realisations consist of a series of constant values, with a different constant value for each realisation. A law of large numbers does not apply on this case, as the limiting value of an average from a single realisation takes the random value determined by $Y$, rather than taking the expected value of $Y$.

I cannot understand the reason for the law of large numbers not being applicable. Any explanation would help.

• Perhaps if you include what exactly are the conditions that must be satisfied in order for your favorite law of large numbers to hold, the answer to your question might be glaringly obvious. Please check carefully that you haven't left out any words beginning with "indepen...." in the conditions that you state, and think as to whether $X_1$ and $X_2$ satisfy this requirement. – Dilip Sarwate Aug 30 '18 at 14:11
• This part is quoted from the Wikipedia article I read, exactly as it was. So, nothing has been left out. I don't understand what you're implying. And I have included the part about the law of large numbers. – Curiosity Aug 30 '18 at 14:13
• Ok, let's try it again from the top. Look in Wikipedia about the laws of large numbers and edit your question to say something like "Now, according to Wikipedia, one version of the law of large numbers states that if ...." and include all the conditions on $X_1, X_2, \ldots$ that this law needs in order to hold. Then think a little about whether $X_1=Y$ and $X_2=Y$ satisfy this condition. – Dilip Sarwate Aug 30 '18 at 14:22

I believe that you are thinking that each value of $X_t$ is determined by a different realisation of $Y$, which in this example is not true.

Suppose that $Y$ is the value that comes out from a dice throw. Thus, $Y$ can achieve any integer value between $1$ and $6$. Throw the dice once. Suppose that you get $2$. Then, according to the definition in the example, your time-series would be:

$$X_t=2 \qquad {\text{for all }}t$$

What I think the quote from Wikipedia is saying is that the mean value of this stochastic process doesn't coincide with the mean value of the random variable $Y$, as

$$\mathbb{E}[X_t]=2 \neq 3.5 = \mathbb{E}[Y]$$

If you constructed the time-series letting each value of $X_t$ be determined by a new throw of the dice, then the means would match.

• Thanks, I got it now. That was a very lucid explanation indeed. – Curiosity Aug 30 '18 at 18:09
• @Curiosity Glad to help! If it solved your doubt, please mark the answer as accepted so it can help future readers. – Tendero Aug 30 '18 at 23:34