# Fundamentals¶

## Odds¶

$O(E) = \frac{P(E)}{P(\bar{E})}$

## Variance and Standard Deviation¶

Spread of probability mass about the mean.

### Variance¶

$Var(X) = E[(X- E[X])^2] = E[X^2] - E[X]^2$

Computation as sum:

$Var(X) = \sum_{i=1}^n p(x_i) (x_i 0 \mu)^2$

### Standard Deviation¶

$\sigma = \sqrt{Var(X)}$

### Standard Error¶

Standard error(SE) of a parameter is the standard deviation(SV) of its sampling distribution or an estimate of the SV. If the parameter or the statistic is the mean, it is called the standard error of the mean(SEM). SEM can be expressed as

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

where n is the sample size.

### Standard Deviation vs. SEM¶

SD quantifies scatter, and SEM how precisely you know the true mean of the population. SEM takes into account both the values of the SD and the sample size. They both use the same units of the data. SEM, by definition, is always smaller than the SD. SEM gets smaller as your sample size, $$n$$, gets larger. This makes sense, because the mean of a large sample is likely to be closer to the true population mean than is mean of a small sample.

## Likelihood¶

A function of the parameters of a statistical model given data.

## Joint distribution¶

A probability distribution for two or more variables. OR A joint distribution is a distribution which is joint.

## Conditional Probabilities¶

$P(A|B) = \frac{A \cap B}{P(B)}$

### Multiplication Rule¶

Reference: Introduction to Probability by Bertsekas pg.24. Assuming that all of the conditioning events have positive probability,

$P(\cap_{i=1}^{n} A_i) = P(A_1)P(A_2|A_1)P(A_3|A_1 \cap A_2) \cdots P(A_n|\cap_{i=1}^{n-1} A_i)$

### Total Probability Theorem¶

Let $$A_1,\cdots ,A_n$$ be disjoint events and assume that $$P(A_i) > 0$$ for all i. Then, for any event B, we have

$\begin{split}\begin{split} P(B) &= P(A_1 \cap B) + \cdots + P(A_n \cap B) \\ &= P(A_1)P(B|A_1)+ \cdots + P(A_n)P(B|A_n) \end{split}\end{split}$

### Conditional distribution¶

For random variables A and B, $$P(A|B)$$ is the probability distribution which describes the change in A when B is changed. Therefore the distribution sums to 1 over all the events in A.

\begin{align}\begin{aligned}A = {a_1, a_2, ... a_n}\\\sum_{i=1}^n P(a_i | B) = 1\end{aligned}\end{align}

### Conditional independence¶

$$X \perp\!\!\!\perp Y | Z$$ denotes that variable X and Y are conditionally independent of each other, given the state of variable Z.

$P_{X,Y |Z}(x,y|z) = P_{X|Z}(x|z)P_{Y|Z}(y|z)$

Intuitively, this means that if we know the value of Z, knowing in addition the value of Y does not provide any information about the value of X. Indeed, provided $$P(y,z) > 0$$, we have

$X \perp\!\!\!\perp Y | Z \Longrightarrow P_{X,|Y,Z}(x|y,z) = P_{X|Z}(x|z)$

Proof:

\begin{split}\begin{align} p(x|y,z) &= \frac{p(x,y,z)}{ p(y,z) } = \frac{p(x,y|z)p(z)}{p(z)p(y|z)} \\ &= \frac{p(x|z)p(y|z)p(z)}{p(z)p(y|z)} = p(x|z)\\ p(x,y,z) &= p(x|y,z)p(y|z)p(z) \quad \text{(General chain rule of probability)} \end{align}\end{split}

### Intuitive examples¶

1. Let $$X_1,X_2,...,X_n$$ denote the cumulative sum of n dice throws, such that $$dom(X_1) = {1,...,6}, dom(X_2) = {2,...,12}$$, etc.
• Is $$X_{n+1}$$ independent of $$X_{n-1}$$? NO.
• Is $$X_{n+1}$$ conditionally independent of $$X_{n-1}$$ given $$X_{n}$$? YES.
1. X =’Location of an airplane now’, Y =’Location of the plane 15s ago’, Z=’Location 15s from now’
• Is Y independent of Z? NO
• Is Y conditionally independent of Z given X? NO

### Practices¶

#### Example 1.¶

Consider the Bayesian network which represents Mr Holmes’ burglary worries as given in the figure: (B)urglar, (A)larm, (W)atson, Mrs (G)ibbon. All vari- ables are binary with states True and False.

< Source: Aalto course CS-E4820: Advanced probabilistic methods >

##### Part a). $$p(B|W)$$¶
$p(B|W) = \frac{p(B,W)}{p(W)}$

Let’s separate the numerator and denominator. Numerator:

\begin{split}\begin{align} p(B,W) &= p(W|B)p(B)\\ p(W|B) &= p(W|A)p(A|B) + p(W|\bar{A})p(\bar{A}|B) = 0.896 \\ p(B,W) &= 0.896 * 0.01 = 0.00896 \end{align}\end{split}

Now denominator:

\begin{split}\begin{align} p(W) &= p(W|B)p(B) + p(W|\bar{B})p(\bar{B})\\ &= 0.52376 \end{align}\end{split}
$p(B|W) = \frac{p(B,W)}{p(W)} = \frac{0.00896}{0.52376} \approx 0.0171$
##### Part b). $$p(B|W,\bar{G})$$¶
$p(B|W,\bar{G}) = \frac{p(B,W,\bar{G})}{p(W,\bar{G})}$

Again, Let’s separate the numerator and denominator. Numerator:

\begin{split}\begin{align} p(B,W,\bar{G}) &= p(W,\bar{G}|B)p(B) \\ p(W,\bar{G}|B) &= p(W|A)p(\bar{G}|A)p(A|B) + p(W|\bar{A})p(\bar{G}|\bar{A})p(\bar{A}|B) \\ &= .2713 \\ p(B,W,\bar{G}) &= 0.002713 \end{align}\end{split}

Now denominator:

\begin{split}\begin{align} p(W,\bar{G}) &= p(W,\bar{G}|B)p(B) + p(W,\bar{G}|\bar{B})p(\bar{B}) \\ p(W,\bar{G}|\bar{B}) &= P(W|A)p(\bar{G}|A)p(A|\bar{B}) \\ &+ P(W|\bar{A})p(\bar{G}|\bar{A})p(\bar{A}|\bar{B})\\ &= 0.3935\\ p(W,\bar{G}) &= 0.002713 + 0.389565 &= 0.392278 \end{align}\end{split}

Therefore,

\begin{split}\begin{align} p(B|W,\bar{G}) &= \frac{p(B,W,\bar{G})}{p(W,\bar{G})} \\ &= \frac{0.002713}{0.392278} \approx 6.916 \times 10^{-3} \end{align}\end{split}

## Marginal distribution¶

Consider a joint probability distribution $$P(\theta 1, \theta 2)$$. A marginal distribution is obtained by integrating over one parameter,

$P(\theta_1) = \int P(\theta_1, \theta_2)d \theta_2$

It gives the probabilities of the variables without reference to the other variables. The contrary of conditional distribution.

For discrete random variables, the marginal probability mass function(PMF) can be written as $$P(X=x)$$.

$P(X=x) = \sum_y P(X=x,Y=y) = \sum_y P(X=x | Y=y)P(Y=y)$

where $$P(X=x,Y=y)$$ is the joint distribution of X and Y.

### Marginal independence¶

Random variable X is marginally independent of random variable Y if, for all $$x_i \in dom(X), y_j \in dom(Y), y_k \in dom(Y),$$,

$\begin{split}P(X=x_i|Y=y_j) = P(X=x_i|Y=y_k) = P(X=x_i) \\ \text{(NOTE: the differences in *j* and *k*)}\end{split}$

That is, knowledge of Y’s value doesn’t affect your belief in the value of X.

## Sample space¶

• Set of all possible outcomes of an experiment
• Size of the set is NOT the sample space
• Outcomes can be sequence of numbers

### Discrete sample space¶

$\begin{split}\text{Discrete = listable} \\\end{split}$

e.g.

\begin{split}\begin{align} {a, b, c} & \quad \text{(finite)} \\ {0, 1, 2, ... } & \quad \text{(infinite)} \end{align}\end{split}

## Independence¶

• Events A & B are independent if $$P(A \cap B) = P(A) \times P(B)$$
• Random variables X and Y are independent if $$F(x, y) = F_X(x) F_Y(y)$$
• Discrete random variables X and Y are independent if $$P(x_i, y_j) = P_X(x_i) P_Y(y_j)$$
• Continuous random variables X and Y are independent if $$f(x, y) = f_X(x) f_Y(y)$$
• $$cov(X, Y) = 0 \iff E[XY] = E[X]E[Y]$$

## Covariance and Correlation¶

The two are very similar. Both describe the degree to which two random variables or sets of random variables tend to deviate from their expected values in similar ways. - Wikipedia

### Covariance¶

Measures the degree to which two random variables vary together, e.g. height and weight of people.

Random variables $$X, Y$$ with means $$\mu_x, \mu_y$$.

$\sigma_{X,Y} cov(X, Y) = E((X - \mu_x)(Y-\mu_y))$

#### Properties¶

• $$cov(aX + b, cY + d) = ac cov(X,Y)$$ for constants $$a,b,c,d$$
• $$cov(X_1 + X_2, Y) = cov(X_1,Y)+cov(X_2,Y)$$
• $$cov(X,X) = Var(X)$$
• $$cov(X,Y) = E(XY) - \mu_x \mu_y$$
• If $$X, Y$$ are independent then $$Cov(X, Y) = 0$$. Warning: The converse is not true, when covariance is 0 the variables might not be independent.

### Correlation¶

It’s like covariance, but it removes the scale. The population correlation coefficient $$\rho_{X,Y}$$ between X and Y is defined by

$\rho_{X,Y} = corr(X, Y) = \frac{cov(X,Y)}{\sigma_X \sigma_Y} = \frac{E[(X - \mu_X)(Y - \mu_Y)]}{\sigma_X \sigma_Y}$

WARNING: It’s not causation.

## Standardization¶

$Y = \frac{X-\mu}{\sigma}$
• $$Y$$ has mean 0 and $$\sigma_Y = 1$$
• Standardizing any normal random variable produces the standard normal.
• If $$X \approx normal$$, then standardized $$X \approx$$ standardized normal
• $$Z$$: standardized normal random variable.

## Random Variables¶

### Random Variable(RV)¶

$X: \Omega \longrightarrow \mathbb{R}$

### Probability Mass Function(PMF)¶

$f_X(x) = P[X = x] = P[{\omega \in \Omega: X(\omega) = x}]$

### Probability Density Function(PDF)¶

$P[a \leq X \leq b] = \int_a^b f(x) dx$

### Cumulative Distribution Function(CDF)¶

$F_X: \mathbb{R} \longrightarrow [0, 1] \quad F_X(x) = P[X \leq x]$

## Exchangability & i.i.d¶

$\begin{split}\text{i.i.d(independently and identically distributed)} \Rightarrow \text{exchangable} \\ \text{, but} \\ \text{i.i.d(independently and identically distributed)} \nLeftarrow \text{exchangable}\end{split}$

Coin tossing is a good example; $$[P(H,H,T) = P(T,H,H)] \Longrightarrow$$ events are independent and their order can be exchanged.

References