ISMT
Bayesian inference and time series analysis
Bayesian inference and time series analysis
[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:
\(Var(aX)=Cov(aX,aX)=E[aXaX]-E[aX]E[aX]\)
\(=a^2E[X^2]-a^2E[X]E[X]=a^2\underbrace{\left[E[x^2]-E[X]E[X]\right]}_{Var(X)}\)
[Probability theory]
Mean of a r.v.?
\(E(X) = \sum_{i=1}^{n} x_i p(x_i)\)
\(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx\)
[Probability theory]
Variance of a r.v.?
\(\text{Var}(X)=E([X-E(X)]^2)\)
\(\text{Var}(X) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)\)
\(\text{Var}(X) = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx\)
[Probability calculation]
Calculate the E[X] of a continuous uniform probability distribution.
A continuous uniform distribution over the interval [a, b] has a probability density function (pdf) given by:
\(f(x) = \frac{1}{b-a} \quad \text{for} \, a \leq x \leq b\)
and \(f(x) = 0\) outside this interval.
Expected Value, E(X), Calculation
The expected value of X for a continuous uniform distribution is calculated using the integral:
\(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx\)
Given the pdf \(f(x) = \frac{1}{b-a} \text{for } a \leq x \leq b\), the integral simplifies to:
\(E(X) = \int_{a}^{b} x \frac{1}{b-a} \, dx\)
because f(x) = 0 outside [a, b].
Now, let's solve this integral:
\(E(X) \\= \frac{1}{b-a} \int_{a}^{b} x \, dx\\= \frac{1}{b-a} \left[ \frac{x^2}{2} \right]_{a}^{b}\\= \frac{1}{b-a} \left( \frac{b^2}{2} - \frac{a^2}{2} \right)\\= \frac{b^2 - a^2}{2(b-a)}\\= \frac{(b-a)(b+a)}{2(b-a)}\\= \frac{b+a}{2}\)
So, the expected value E(X) of a continuous uniform distribution over the interval [a, b] is the midpoint of the interval, which is:
\(E(X) = \frac{a + b}{2}\)
This result shows that for a continuous uniform distribution, the expected value is simply the average of the lower and upper bounds of the distribution's interval.
Let X and Y be two random variables. The conditional expectation of X given Y = y is defined as?
\(E[X|Y = y] = \sum_{x} x \, p_{X|Y}(x|y)\)
\(E[X|Y = y] = \int_{-\infty}^{+\infty} x \, f_{X|Y}(x|y) \, dx\)
[Probability theory]
Let X and Y be two random variables. Assuming expectations exist, what are helpful formulas for the expected value of X and the variance of X?
\(E[X]=E[E[X|Y]]\)
\(Var(X)=Var(E[X|Y])+E[Var(X|Y)]\)
[Statistics]
Let X and Y be two random variables. Then h(X) that minimizes
\(E\left[(Y-h(X))^2 \right]\)
is given by?
\(h(X)=E[Y|X]\)
provided the expectations exist. "The expected value is always the best (linear?) predictor."
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Lernkarten
[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:
\(Var(aX)=Cov(aX,aX)=E[aXaX]-E[aX]E[aX]\)
\(=a^2E[X^2]-a^2E[X]E[X]=a^2\underbrace{\left[E[x^2]-E[X]E[X]\right]}_{Var(X)}\)
[Probability theory]
Mean of a r.v.?
\(E(X) = \sum_{i=1}^{n} x_i p(x_i)\)
\(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx\)
[Probability theory]
Variance of a r.v.?
\(\text{Var}(X)=E([X-E(X)]^2)\)
\(\text{Var}(X) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)\)
\(\text{Var}(X) = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx\)
[Probability calculation]
Calculate the E[X] of a continuous uniform probability distribution.
A continuous uniform distribution over the interval [a, b] has a probability density function (pdf) given by:
\(f(x) = \frac{1}{b-a} \quad \text{for} \, a \leq x \leq b\)
and \(f(x) = 0\) outside this interval.
Expected Value, E(X), Calculation
The expected value of X for a continuous uniform distribution is calculated using the integral:
\(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx\)
Given the pdf \(f(x) = \frac{1}{b-a} \text{for } a \leq x \leq b\), the integral simplifies to:
\(E(X) = \int_{a}^{b} x \frac{1}{b-a} \, dx\)
because f(x) = 0 outside [a, b].
Now, let's solve this integral:
\(E(X) \\= \frac{1}{b-a} \int_{a}^{b} x \, dx\\= \frac{1}{b-a} \left[ \frac{x^2}{2} \right]_{a}^{b}\\= \frac{1}{b-a} \left( \frac{b^2}{2} - \frac{a^2}{2} \right)\\= \frac{b^2 - a^2}{2(b-a)}\\= \frac{(b-a)(b+a)}{2(b-a)}\\= \frac{b+a}{2}\)
So, the expected value E(X) of a continuous uniform distribution over the interval [a, b] is the midpoint of the interval, which is:
\(E(X) = \frac{a + b}{2}\)
This result shows that for a continuous uniform distribution, the expected value is simply the average of the lower and upper bounds of the distribution's interval.
Let X and Y be two random variables. The conditional expectation of X given Y = y is defined as?
\(E[X|Y = y] = \sum_{x} x \, p_{X|Y}(x|y)\)
\(E[X|Y = y] = \int_{-\infty}^{+\infty} x \, f_{X|Y}(x|y) \, dx\)
[Probability theory]
Let X and Y be two random variables. Assuming expectations exist, what are helpful formulas for the expected value of X and the variance of X?
\(E[X]=E[E[X|Y]]\)
\(Var(X)=Var(E[X|Y])+E[Var(X|Y)]\)
[Statistics]
Let X and Y be two random variables. Then h(X) that minimizes
\(E\left[(Y-h(X))^2 \right]\)
is given by?
\(h(X)=E[Y|X]\)
provided the expectations exist. "The expected value is always the best (linear?) predictor."
