# How is $P[X_t\le x_t | X_1,\ldots, X_{t-1}]=P[X_t\le x_t]$ when $X_t\sim WN(0,\sigma^2)$?

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 ?

• 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. Apr 1, 2015 at 12:11

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

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.

• Got the point of $P[X_t\le x_t | X_1,\ldots, X_{t-1}]=P[X_t\le x_t]$ . (+1)
– time
Apr 1, 2015 at 12:01
• But didn't understand why is it not interesting for forecasting ?
– time
Apr 1, 2015 at 12:04
• 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. Apr 1, 2015 at 12:11