Sequences of Random Variables

March 2, 2025 · 2 min read · Page View:

If you have any questions, feel free to comment below.

Random vector #

Substitute certain variables in $F(X.……,)y$ by $\infty$, we get the joint distribution of the remaining variables:

$F\left(x_{1}, x_{3}\right)=F\left(x_{1}, \infty, x_{3}, \infty\right)$

$f\left(x_{1}, x_{3}\right)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f\left(x_{1}, x_{2}, x_{3}, x_{4}\right) d x_{2} d x_{4}$

Example - Conditional dist. of uniform r.v. on diamond

Random matrix #

Mean and Variance of Random Vector #

The Mx means the mean vector of X, which can be expressed as:

eg.

But if the two events are independent, then: Cov(X,Y) = 0 In the variance matrix there only will be Var(X) and Var(Y)

Functions of Random Vectors #

Density: Jacobin matrix #

Independence & IID R.V.s #

idd rvs

Order Statistics #

Stochastic convergence: different modes #

for a specific e, X(e) is a sequence of numbers that might or might not converge to a limit.

eg. X_n = n / (n + 1), n = 0, 1, 2, … The sequence converges to 1 when n to infinity.

which means in any neighborhood of 1,

∀ ε > 0, ∃ n_0(ε) s.t. for n ≥ n_0(ε) |X_n − 1| ≤ ε.

We can pick the $\epsilon$ to be very small, make sure the sequence will be trapped after reaching n_0(ε).

And as $\epsilon$ decreases, the n_0(ε) will increase. eg. when $\epsilon = 0.1$, n_0(ε) = 10; when $\epsilon = 0.01$, n_0(ε) = 100; when $\epsilon = 0.001$, n_0(ε) = 1001;

Convergence everywhere #

e. $X_n$ to $X$ as $n$ to $\infty$

a.e. $P(X_n$ to $X) = 1$ as $n$ to $\infty$

m.s. $E(|X_n - X|^2)$ to $0$ as $n$ to $\infty$

p. $P(|X_n - X| > \epsilon)$ to $0$ as $n$ to $\infty$

d. Fn(x) to F(x) as n to infinity

Limit Theorems #

Let $S_{n}=X_{1}+\cdots+X_{n}$ be the sum of the first n measurements, the X1, X2.. of iid with mean $\mu$ and variance $\sigma^2$.

Because of independence,

$$ var\left(S_{n}\right)=var\left(X_{1}\right)+\cdots+var\left(X_{n}\right)=n \sigma^{2} $$

With n increasing, the dist of S will spread out.

Direct research on $S_n$ cannot be done, but we can study the sample mean $M_n$.

$$ M_{n}=\frac{S_{n}}{n}=\frac{X_{1}+\cdots+X_{n}}{n} $$

The law of large numbers #

As we disused before $E[M_{n}]=\mu_{n}$ var $(M_{n})=\frac{\sigma^{2}}{n}$

The variance of $M_n$ decreases to 0 as $n$ increases. And $M_n$ converges to $\mu$ as $n$ increases., which means the sample mean will converge to the real mean $\mu$.

Markov and Chebyshev Inequalities #

And if the sigma is unknown, for x in [a, b], we may use the bound (b-a)^2/4 as the substitute. for all c > 0.

Applications #

Weak Law of Large Numbers #

Central Limit Theorem #

The normaltive average defined as $Z_{n}=\frac{S_{n}-n \mu}{\sigma \sqrt{n}}$

1.subtract nu from $S_{n}$ to obtain the zero-mean random variable $S_{n}-n \mu$
2.divide the result by $\sigma \sqrt{n}$ to get the normalized variable $Z_n$ Recall Var(Sn) = $n\sigma^2$

It can be seen that $E[Z_{n}]=0$ var $$[Z_{n}] = 1$$

So when n increases to infinity, the distribution of $Z_n$ will be close to the standard normal distribution.

So the mean or var will independent of n.

The approximations based on ctl #

Let $S_{n}=X_{1}+\cdots+X_{n}$ where the X are iid.random variables with mean u, variance $\sigma^2$

If n is large, the probability $P(S_{n} ≤c)$ can be approximated by treating $S_{n}$ as if it were normal, according to the following procedure. Pay attention to the c!!!

calculate the mean $n\mu$ and variance $n\sigma^2$
calculate the normalized variable $Z_n = \frac{S_n - n\mu}{\sigma\sqrt{n}}$
use the approximation $$P(S_{n} ≤c) ≈\Phi(z)$$