ISMT
Bayesian inference and time series analysis
Bayesian inference and time series analysis
Kartei Details
| Karten | 14 |
|---|---|
| Sprache | English |
| Kategorie | Mathematik |
| Stufe | Universität |
| Erstellt / Aktualisiert | 02.10.2022 / 05.02.2024 |
| Lizenzierung | Kein Urheberrechtsschutz (CC0) |
| Weblink |
https://card2brain.ch/box/20221002_ismte136
|
| Einbinden |
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|
[Statistics]
Given
\(E[X]=\sum_i x_i f(x_i)\)
\(E[X^2] = \text{?}\)
\(E[X^2]=Var[X]+E[X]^2\)
We can see that \(E[X^2] \neq E[X]^2 \text{ if } Var[X]\neq 0\)
[Time series analysis]
If AR(1) is causal and given as
\(x_t=\phi x_{t-1}+w_t\text{, where }w_t\sim wn(0, \sigma_w^2)\)
\(Var[x_t]= \text{?}\)
\(\begin{align} &Var[x_t]=var(\alpha + \phi x_{t-1}+w_t) \\&= 0 + var(\phi x_{t-1}) + var(w_t) \\&= \phi^2var(x_{t-1})+\sigma_{w}^2 \\&= \phi^2var(x_{t})+\sigma_{w}^2 &\text{| as series is stationary} \\&= \frac{\sigma{_w}^2}{1-\phi^2}\end{align} \)
[Statistics]
\(cov(aX + bY, cW + dV) = \text{?}\)
\(cov(aX + bY, cW + dV) = ac \cdot cov(X, W) + ad \cdot cov(X, V) + bc\cdot cov(Y, W) + bd\cdot cov(Y, V)\)
[Statistics]
\(cov(X, Y) = \text{?}\)
\(cov(X, Y)=E[XY]-E[X]E[Y]\)
[Time series analysis]
If AR(1) is causal and given as
\(x_t=\phi x_{t-1}+w_t\text{, where }w_t\sim wn(0, \sigma_w^2)\)
(a) Stationary solution?
(b) \(E(x_t)= \text{?}\)
(c) \(\gamma(h)= \text{?}\)
(d) \(\rho(h)= \text{?}\)
[Time series analysis]
If MA(1) is given as
\(x_t=\theta w_{t-1}+w_t\text{, where }w_t\sim wn(0, \sigma_w^2)\)
(a) \(E(x_t) = \text{?}\)
(b) \(\gamma(h)= \text{?}\)
(c) \(\rho(h)= \text{?}\)
[Time series]
In general, the correlation of any (stationary) time series can be calculated through...?
\(\rho(h)=\frac{\gamma(h)}{\gamma(0)}\)
Note:
- \(\gamma(0)\) is the variance of the series
- This only works due to stationarity. Pearson correlation coefficient is actually \(\frac{cov(X, Y)}{\sigma_X\sigma_Y}\)
[Statistics]
\(Var(aX) = \text{?}\)
\(Var(aX) = a^2Var(X)\)
Easy proof:
\(Var(aX)=Cov(aX,aX)=E[aXaX]-E[aX]E[aX]\)
\(=a^2E[X^2]-a^2E[X]E[X]=a^2\underbrace{\left[E[x^2]-E[X]E[X]\right]}_{Var(X)}\)
