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In this slide , p.30 , p.31 , it is written that :

White noise : $X_t\sim WN(0,\sigma^2)$ i.e., ${\{X_t}\}$ uncorrelated, $\mathbb E[X_t]=0, \mathbb V[X_t] =\sigma^2$ Example : i.i.d noise : ${\{X_t}\}$ independent and identically distributed $$P[X_1\le x_1,\ldots X_t\le x_t]=P[X_1\le x_1]\ldots P[X_t\le x_t]$$ Not interesting for forecasting : $$P[X_t\le x_t | X_1,\ldots, X_{t-1}]=P[X_t\le x_t]$$ $$P[X_t\le x_t]=\phi(x_t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x_t}\exp[-\frac{x^2}{2}]dx$$

But I have not understood :

  • How is $P[X_t\le x_t | X_1,\ldots, X_{t-1}]=P[X_t\le x_t]$? Why is it not interesting for forecasting?

  • If $X_t\sim WN(0,\sigma^2)$ , then isn't $$P[X_t\le x_t]=\phi(x_t)=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{x_t}\exp[-\frac{x^2}{2\sigma^2}]dx?$$

But in the slide $\sigma^2$ is not considered . Why?

  • $X_t\sim WN(0,\sigma^2)$

i.e., ${\{X_t}\}$ uncorrelated. If ${\{X_t}\}$ were correlated what would be the structure ?

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    $\begingroup$ It is not interesting for forecasting because the variable at some given time is independent of its history. This makes it quite hard to forecast given the history. $\endgroup$
    – hejseb
    Apr 1, 2015 at 12:11

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I hope to answer at least one of your questions:

$$P[X_t\le x_t | X_1,\ldots, X_{t-1}]=P[X_t\le x_t]$$

follows from iid-ness, where the first "i" stands for independent. By probability rules, $P(A|B)=P(A)$ if $A$ and $B$ are independent.

As for why that is not interesting for forecasting, my guess is that the joint probability stated before is not interesting for that purpose.

And I guess the example simply sets $\sigma^2=1$. These are just slides, often not meant to be self-explanatory.

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    $\begingroup$ Got the point of $P[X_t\le x_t | X_1,\ldots, X_{t-1}]=P[X_t\le x_t]$ . (+1) $\endgroup$
    – time
    Apr 1, 2015 at 12:01
  • $\begingroup$ But didn't understand why is it not interesting for forecasting ? $\endgroup$
    – time
    Apr 1, 2015 at 12:04
  • $\begingroup$ Because in forecasting you usually answer questions like "given that you know A (e.g. present and past) what do you expect for B (e.g. the future)" and that question is naturally couched in terms of conditional probabilities, not joint ones. $\endgroup$ Apr 1, 2015 at 12:11

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