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CSCI 328 - Formulas & Identities Cheatsheet

This cheatsheet captures all major formulas and identities from the course, organized by topic for quick reference during problem-solving and exam preparation.


Pr(Ac)=1Pr(A)\Pr(A^c) = 1 - \Pr(A)

Two events AA and BB are independent if:

Pr(AB)=Pr(A)Pr(B)\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)

For any random variables X1,X2,,XnX_1, X_2, \ldots, X_n (with or without independence):

E[i=1nXi]=i=1nE[Xi]E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n E[X_i] Pr(A1A2An)i=1nPr(Ai)\Pr(A_1 \cup A_2 \cup \cdots \cup A_n) \leq \sum_{i=1}^n \Pr(A_i)

For an event AA:

XA={1if A occurs0otherwise,E[XA]=Pr(A)X_A = \begin{cases} 1 & \text{if } A \text{ occurs} \\ 0 & \text{otherwise} \end{cases}, \quad E[X_A] = \Pr(A)

XBernoulli(p)X \sim \text{Bernoulli}(p):

E[X]=p,Var(X)=p(1p)=pq(where q=1p)E[X] = p, \quad \text{Var}(X) = p(1-p) = p \cdot q \quad \text{(where } q = 1-p\text{)}

XBinomial(n,p)X \sim \text{Binomial}(n, p) (sum of nn independent Bernoulli trials):

E[X]=np,Var(X)=np(1p)=npqE[X] = np, \quad \text{Var}(X) = np(1-p) = npq Pr(X=k)=(nk)pk(1p)nk\Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

XGeometric(p)X \sim \text{Geometric}(p) (number of trials until first success):

E[X]=1p,Var(X)=1pp2E[X] = \frac{1}{p}, \quad \text{Var}(X) = \frac{1-p}{p^2}

Expected number of items needed to collect all nn distinct types (with uniform sampling):

E[# items]=n(1+12+13++1n)=nHnnlnnE[\text{\# items}] = n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right) = n H_n \approx n \ln n

where HnH_n is the nn-th harmonic number.


Var(X)=E[X2](E[X])2\text{Var}(X) = E[X^2] - (E[X])^2

If X1,X2,,XnX_1, X_2, \ldots, X_n are independent:

Var(i=1nXi)=i=1nVar(Xi)\text{Var}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \text{Var}(X_i) σ=Var(X)\sigma = \sqrt{\text{Var}(X)}

For any non-negative random variable XX and t>0t > 0:

Pr(Xt)E[X]t\Pr(X \geq t) \leq \frac{E[X]}{t}

Intuition: Probability of extreme deviation is bounded by expected value ratio.

For any random variable XX and t>0t > 0:

Pr(XE[X]t)Var(X)t2\Pr(|X - E[X]| \geq t) \leq \frac{\text{Var}(X)}{t^2}

Alternatively:

Pr(Xt)Var(X)t2\Pr(X \geq t) \leq \frac{\text{Var}(X)}{t^2}

Intuition: Values far from the mean are unlikely if variance is small.

For X=i=1nXiX = \sum_{i=1}^n X_i where XiX_i are independent Bernoulli random variables, and μ=E[X]\mu = E[X]:

(1) General Form — For any δ>0\delta > 0:

Pr(X(1+δ)μ)(eδ(1+δ)1+δ)μ\Pr(X \geq (1+\delta)\mu) \leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu

(2) Simplified Form — For 0<δ10 < \delta \leq 1:

Pr(X(1+δ)μ)eμδ2/3\Pr(X \geq (1+\delta)\mu) \leq e^{-\mu\delta^2/3}

(3) For Large Deviations — For R6μR \geq 6\mu:

Pr(XR)2R\Pr(X \geq R) \leq 2^{-R}

Intuition: Exponential bounds for sums of independent Bernoulli variables.


PART VI: BIRTHDAY PARADOX & COLLISION PROBABILITY

Section titled “PART VI: BIRTHDAY PARADOX & COLLISION PROBABILITY”

Probability that mm balls thrown into nn bins have no collisions:

Pr(no collision)=nnn1nn2nnm+1n\Pr(\text{no collision}) = \frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{n-m+1}{n}

This can be written as:

Pr(no collision)=i=0m1(1in)\Pr(\text{no collision}) = \prod_{i=0}^{m-1} \left(1 - \frac{i}{n}\right)

Approximate Collision Probability (for m2nm^2 \ll n)

Section titled “Approximate Collision Probability (for m2≪nm^2 \ll nm2≪n)”
Pr(collision)1em2/(2n)\Pr(\text{collision}) \approx 1 - e^{-m^2/(2n)}

Birthday Paradox (specific case: n=365n = 365, m=23m = 23)

Section titled “Birthday Paradox (specific case: n=365n = 365n=365, m=23m = 23m=23)”

Probability that 23 people have a shared birthday exceeds 50%.


When nn balls are thrown into nn bins:

Pr(bin is empty)=(11/n)ne1\Pr(\text{bin is empty}) = (1 - 1/n)^n \approx e^{-1} E[# empty bins]=ne10.368nE[\text{\# empty bins}] = n \cdot e^{-1} \approx 0.368n

When nn balls are thrown into nn bins:

E[max load]=Θ(lnnlnlnn)E[\text{max load}] = \Theta\left(\frac{\ln n}{\ln \ln n}\right) Pr(max load>3lnnlnlnn)1n\Pr\left(\text{max load} > \frac{3\ln n}{\ln \ln n}\right) \leq \frac{1}{n}