Prob Stat

Notes for reviewing my probability and statistics course at Northeastern.

Sample Spaces

  • Experiment :: Repeatable procedure with a set of possible results
  • Sample Outcome :: One of many possible results of an experiment
  • Sample Space :: S$: The set of all possible outcomes of an experiment
  • Event :: A \subset S$ : 0 or more outcomes of an experiment
  • Probability :: P(A) = \frac{n(A)}{n(S)}$

Set Theory

  • Discrete Set :: A finite or countable set. i.e. tossing a coin until a head is received: S = \{ H, TH, TTH, \dots \}$
  • Continuous Set :: A range of values. I.e. getting a number smaller than 1 from within a real number between 0 and \sqrt{2}$: $S = [0, \sqrt{2}]$, $A = [0, 1)$
  • Intersection :: A \cap B = \{x | x \in A\ and\ x \in B\}$ for some events A, B

+ Disjoint :: A \cap B = \emptyset$. Also known as "mutually exclusive"

  • Union :: A \cup B = \{x | x \in A\ or\ x \in B\}$
  • Complement :: A^{c} = \{x \in S | x \not{\in} A\}$

DeMorgan's Law(s)

1. (A \cup B)^{c} = A^{c} \cap B^{c}$

1. (A \cap B)^{c} = A^{c} \cup B^{c}$

Probability Function

P$ assigns a real number to any event of a sample space, and follows the following axioms for a sample space that's finite:

  • P(A) \geq 0$ for all $A$
  • P(S) = 1$
  • A \cap B = \emptyset \implies P(A \cup B) = P(A) + P(B)$, or $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ otherwise
  • P(\cup_{i=1}^{\infty} A_{i}) = \sum_{i=1}^{\infty} P(A_{i})$ if any $A_1, A_2, A_3 \dots$ are mutually exclusive in $S$

Conditional Probability

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

P(A \cap B) = P(B|A)P(A) = P(A|B)P(B)$

Solving conditional probability: Make a tree! Fork at each choice, recording the probability on each "leaf". Then use the leaves to assess a series of events.


A$ and $B$ are _independent_ if $P(A \cap B) = P(A)P(B)$; put another way, $P(A|B) = P(A) \iff P(B|A) = P(B)$

For more than two sets:

  • P(A \cap B \cap C) = P(A)P(B)P(C)$
  • P(A \cap B) = P(A)P(B)$, $P(A \cap C) = P(A)P(C)$, $P(B \cap C) = P(B)P(C)$ and so on...



S_{n} = \sum_{k=0}^{n}r^{k} = 1 + r + r^{2} + \dots + r^{n}$

For -1 < r < 1$, the sum converges:

S \equiv S_{\infty} = \sum_{k=0}^{\infty}r^{k} = \frac{1}{1 - r}$, for $|r| < 1$

  • [PDF] :: p_{Y}(k) = (1 - p)^{k - 1}p$
  • mean :: \frac{1}{p}$

Probability Distributions

(^{n}_{r}) = _{n}C_{r} = \frac{n!}{(n - r)! r!}$

Binomial Distribution

Given n _independent_ trials with two outcomes and a constant P(success) for each outcome, P(k) = (^{n}_{k}) * p^{k}(1 - p)^{n - k}$ for $k = 0, 1, 2, \dots, n$.

  • [PDF] :: P(X = k) = (^{n}_{k}) * p^{k}(1 - p)^{n - k}$

+ calculate: use ~binompdf~

  • [CDF] :: P(X \leq t) = \sum_{k=0}^{t}(^{n}_{k}) * p^{k}(1 - p)^{n - k}$

+ calculate: use ~binomcdf~

Bernoulli Distribution

Essentially a [Binomial Distribution] where n = 1$.

P(k) = (^{1}_{k}) * p^{k}(1 - p)^{1 - k}$

Hypergeometric Distribution

P(x) = \frac{(^{k}_{x})(^{N-k}_{n-x})}{(^{N}_{n})}$

  • Random selection of n$ items without replacement from a set of $N$ items
  • _Not_ guaranteed P(success) stays constant.
  • k$ items are success; $N - k$ items are failures.

This can be considered a generalization of the [Binomial Distribution].

Uniform Distribution

All values in a range are equally likely.

For some interval [a, b]$:

  • [PDF] :: \frac{1}{b - a}$
  • [CDF] :: \frac{x - a}{b - a}$

Exponential Distribution

  • [PDF] :: f_{X}(x) = \lambda e^{-\lambda x}$ for $x \geq 0$
  • [CDF] :: F_{X}(x) = 1 - e^{-\lambda x}$
  • mean :: E(X) = \frac{1}{\lambda}$
  • var :: Var(X) = \frac{1}{\lambda^{2}}$

Poisson Distribution

Counts the number of occurences per unit of measurement - over a specific period of time, a specific area, or volume, etc...

The probability of an event occuring in a unit of measurement must be the same for all similar units;

for example, if the unit of measurement is a month, then the probability must be the same for all months.

Poisson is often used to approximate the binomial distribution: given that n$ is large ($n \geq 100$) and $p$ is small, we can let $\lambda = np$!

In other words, \lambda := np$ -> (average number of occurences per unit) * (length of observation period)

Use ~poissonpdf~ and ~poissoncdf~ calculator unctions to approximate this.

  • [PDF] :: p_{X}(k) = P(X = k) := \frac{\lambda^{k}e^{-\lambda}}{k!}$
  • mean :: E(X) = \lambda$
  • variance :: Var(X) = \lambda$

Normal Distribution

Density Functions

  • Discrete Random Variable :: Some X$ such that:

+ X: S \rightarrow \mathbb{R}$

+ X$ is a countable subset of $\mathbb{R}$

+ Motivation: Constrain the sample space to a smaller sample space, using a single variable to represent each outcome we're investigating. If we're looking at pairs of numbers, for example, we only care that the sum of the pair is the same, so we consider (1, 2) and (2, 1) to be the same outcome.

  • Continuous random variable :: Like discrete,

but ranges over a continuous interval of \mathbb{R}$ instead.

Two continuous random variables are _independent_ if some functions, g(x)$ and $h(x)$, exist such that:

- f_{X,Y}(x,y) = g(x)h(y)$

- f_{X}(x) = g(x)$

- f_{Y}(y) = h(y)$

In other words, one should be able to multiply the results of the marginal pdfs to produce the joint pdf for the two variables, and vice versa.


(Probability Density Function)


For every X$, a probability density function (pdf) looks like:

p_{x}(k) = P(X = k) := P(\{s \in S | X(s) = k\})$, where $p_{x}(k): \mathbb{R} \rightarrow \mathbb{R}$.

Here, s$ and $S$ are from the original sample space we're sampling from.


Some f_{x}(x)$ satisfying:

1. f_{x}(x) \geq 0$

2. \int_{-\infty}^{\infty}f_{x}(x) = 1$

P(a \leq X \leq b) = \int_{b}^{a}f_{x}(x)dx$

Joint PDF

p_{X,Y}(x,y) := P(X = x, Y = y)$, satisfying:

1. p_{X,Y}(x,y) \geq 0$

2. \sum_{all y} \sum_{all y} p_{X,Y}(x,y) = 1$


Given the joint pdf of X$ and $Y$, the _marginal_ pdfs of $X$ and $Y$ are:

  • p_{X}(x) = \sum_{all y}p_{X,Y}(x,y)$
  • p_{Y}(y) = \sum_{all x}p_{X,Y}(x,y)$

P((X,Y) \in R) = \int \int_{R}f_{X,Y}(x,y) dx dy$

When solving - identify the bound that's dependent on the other!

It can really help to plot out some 2D plane, then graph the relationship between the two continuous random variables.

From this graph it's often fairly easy to identify, and thus estimate, the area we're investigating; this guides us to investigate what we should learn more from!

Absolutely worth working through some pracice problems.


Cumulative Distribution Function


F_{x}(t) = P(X \leq t) := P(\{s \in S | X(s) \leq t\})$, where $F_{x}(t): \mathbb{R} \rightarrow \mathbb{R}$.

Generally, F_{x}(t) = \int_{-\infty}^{t}p_{x}(t)$; the pdf represents the probability of the discrete random variable being a specific value, while the cdf represents the probability of all outcomes occuring less than some outcome upper bound $t$.


F_{x}(x) = P(X \leq x) = \int_{-\infty}^{x}p_{x}(x)dx$

P(a \leq X \leq b) = F_{X}(b) - F_{X}(a)$

Expected Value

A generalization of the concept of "average". The name's on the tin - it's a value that represents the proportionally weighted, expected result.

For example, if I have a 5% chance at 100 and a 95% chance at $20, the expected value would be $100 * 0.05 + 0.95 * 20$, so $24.


E(X) = \sum_{all \ k} k * p_{X}(k)$


E(X) = \int_{-\infty}^{\infty} x * p_{X}(x) dx$

Other Properties

E(aX + bY) = aE(X) + bE(Y)$ for any random varaibles $X, Y$ and numbers $a$ and $b$.

As such, if X$ and $Y$ are _independent_, then $E(XY) = E(X)E(Y)$.

Sample Mean

Denoted as \bar{X}$

\bar{X} = \frac{1}{n}(X_{1} + X_{2} + \dots + X_{n})$



"Middle number" of the distribution, or the average of the two middle numbers if the cardinality is even; the standard definition.


m$ such that $\int_{-\infty}^{m}f_{Y}(y) dy = 0.5$.

Finding the median:

1. Integrate and substitute.

2. Factor in terms of and solve for m$.


A measure of how far the distribution spreads from its mean.

Var(X) := E((X - \mu)^{2})$, where:

  • \mu = E(X)$ is the mean of $X$
  • \sigma := \sqrt{Var(X)}$ is the standard deviation of $X$

If X$ and $Y$ are independent, then:

Var(aX + bY) = a^{2}Var(X) + b^{2}Var(Y)$

In general:

Var(aX + bY) = a^{2}Var(X) + b^{2}Var(Y) - 2abCov(X,Y)$


where Cov(X, Y)$, the _covariance_ of X and Y, is:

Cov(X,Y) := E(XY) - E(X)E(Y)$

As can be assumed, if X$ and $Y$ are independent, then $Cov(X,Y) = 0$.


Corr(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$

A measurement of _correlation_; if positive, then the two random variables increase together;

if negative, one increases while the other decreasese and vice versa.

Double Integrals

Fubini's Theorem: \int_{a}^{b} \int_{c}^{d} f(x,y) dy dx = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy$, given $a \leq x \leq b$ and $c \leq y \leq d$

To identify the region in \mathbb{R}^{2}$ to integrate over, use the inside first.

Treat the unevaluated integral variable as a constant and just integrate with respect to a constant.

Often one variable is much easier to integrate than the other; pick the right one to use! This takes practice.

TODO Problems to Practice


Graphing PDF, CDF

Converting between the two, esp. with continuous piecewise

what's with that variance theorm and E(g(X))? practice those problems.

there are some good exercises for deriving expected value and variance available in the textbook - review these!

do the exam




5.x, especially the problem i missed on the last exam

exam problems and how exactly to approach them

the rest of the 6s and 7s, though i can probably wing those just

Types of Problems

Exam 1

Probabilities and Sets

i.e. P(A), P(A \cup B), P(A \cap B^{c})$ -> find something

Draw things out as Venn diagrams to help visualize

Nice properties:

  • P(A \cup B) = P(A \cap B^{c}) + P(A \cap B) + P(A^{c} \cap B)$
  • Bayes: P(A|B) = \frac{P(A \cap B)}{P(B)}$

Simple Probabilities

Generally, use some arrangement of nCr$ ;

P(at most x) = 1 - P(at least x); vice vers

Conditional Probabilities with Scenarios

Draw a tree:

  • At the root, no branch is chosen
  • After the root, choose what info you have more about: i.e. if you're looking for P(Defective | built at plant X), draw a tree with the first set of descendant notes representing the plant at which the thing was constructed, then from there the chances that the product was defective _given_ the plant branching off of each.

From here, can use Bayes rule to fill out the tree, then find desiresd probability.

Exam 2

distributions to know: binomial,

Finding Mean, stdev, sample from given problem

u = np$

\sigma = \sqrt{np(1-p)}$

P(X=a) = nCa * P(a)^{a}(1-P(a))^{n-a}$ ~ ~binompdf~ w/ n, P(a), a

Finding P in range: P(a < X \leq b) = cdf(500, b, P(x)) - cdf(500, a, P(x))$ gives CDF for that range!

Find cdf of discrete set of scenarios

1. Write out each possible scenario and associated value of random variable

2. Use P(scenario) * (num occurences of scenario) for each random variable value to compute some P(X=res) for each

3. Assemble into table; P(X=k) for values x=1, x=2, x=3 for example.



when converting to cdf, outside the range provided for the pdf, must state that the value of the cdf is 0 before and 1 following the range; otherwise the cdf won't function outside of it, but the range for a cdf should support anything



Find E, Var given density function

1. E(X) = \int_{a}^{b} x f_{X}(x) dx$ over provided range $a \leq f_{X}(X) \leq b$

2. Var(X) = E(X^{2}) - (E(X))^{2}$

a. where E(X^{2}) = \int_{a}^{b} x^{2} f_{X}(x) dx$

Piecewise CDF->PDF, PDF -> CDF

derivative, integral respectively of each function provided over each range

Joint PDF, CDF

1. Find marginal pdfs for each; for x, integrate by dy, and for y integrate by dx

2. Set up integral for E(XY)$; this should be some $\int_{c}^{d} \int_{a}^{b} x*y*f_{XY}(x,y) dx dy$ for bounds $a \leq x \leq b$ and $c \leq y \leq d$ . If bounds overlap, reference the relationship between them (i.e. $0 < y < x < 1 \implies a=y, b=1, c=0, d=1$), as the bounds of one are dependent on the bounds for the other.

3. Find the [Covariance] Cov(X,Y) := E(XY) - E(X)E(Y)$. Use the marginal PDFs and integrate to find corresponding $E$, then follow the formula

finding c for some pdf with constant to solve for

TODO Combining Random Variables

i.e. test 2: 13, 14

Exam 3

use normal distribution with continuity correction to estimate probability in bounds

1. find mean (\mu$) and standard deviation ($\sigma$) of the provided scenario for the sample mean; note that $\sigma(\bar{x}) = \frac{\sigma}{\sqrt{n}}$

2. state that the normal approximation can be used if given "normal approximation" or n \geq 30$; "using Central Limit Theorem" if using this approximation, where $\bar{X} \tilde{=} N(\mu, \frac{\sigma}{\sqrt{n}})$

a. TODO when to use \frac{\sigma^{2}}{n}$?

3. continuity correction: "round up or down" to the nearest 0.5 so that the interval encapsulates the intended population. i.e. if interval is \leq$, it's necessary to ensure interval will encapsulate upper and lower bound

4. apply normalcdf: normalcdf(lower, upper, \mu, \sigma)$; where with sample mean, use $\sigma(\bar{x}) = \frac{\sigma}{\sqrt{n}}$ instead. Use $10^{99}$ to replace upper and lower bounds (negative for low) as needed to fill in provided open P(..) intervals.

solve for interval given resultant probability

1. Investigate the interval; sketch it out relative to a normal distribution. If it's two tailed, mark that

2. Finding the bound asked for: invNorm(area before interval, \bar{x}, \sigma(\bar{x}))$ provides such a bound.

probabilities with poisson dist.


finding confidence interval

maximum likelihood estimation

1. Find L(\theta) = \prod_{i=1}^{n}\f_{X}(x)$

Exam 4

unbiasted estimators variance

Some \hat{\theta}$ is an unbiased estimator for $\theta$ if $E(\hat{\theta}) = \theta$, so:

1. Set up variance formula: ignore constants, and take variance of the random variables used to calculate the new random variable

2. Substitute based on what's provided for these existing random variables; i.e. if E(X) = E(Y) = \theta$, can substitute the expected value of each there for theta when calculating

3. if original value is reached after evaluating, then it's an unbiased estimator!


examine the calculation for the random variable, squaring the constants and taking the variance of the random variables used to calculate it

find convidence interval

T test

1. State facts: S, df, \alpha$

2. calculate t_{\frac{\alpha}{2}, df}$ with $invT(1 - \alpha, df)$.

3. Find interval: \bar{X} = t_{\frac{\alpha}{2}, df} \pm \frac{s}{\sqrt{n}}$; can use TInterval

Z test

statistical test

critical values and errors

1. find critical value in nonstandard form: typically invNorm(accept-area, average, \frac{\sigma}{\sqrt{n}})$

2. Find type 1 error: typically \alpha$

3. find type 2 error (given that real mean, \mu(H_{a})$): $\beta = P(TypeII) = P(accept H_{0} | \mu = H_{a})$ -> $normalcdf(b1, b2, \mu(H_{a}), \sigma(\bar{x}))$

4. Find the power of the test: Power = 1 - \beta$

Additional Material

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