σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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Probability skill›Module 1 · Events and conditioning›Topic 1.1·~35 minBeginner

Sample spaces and events

Skill positionModule 1 of 5 · Topic 1 of 2
1.1 Sample spaces and events
—Lab score— / 100Total XP—Streak—

Outcomes, events and probabilities

A random experiment has a sample space Ω\OmegaΩ, the set of all possible outcomes: the six faces of a die, or whether a stock closes up, down or flat. An event is a set of outcomes you care about, such as “”. A probability PPP assigns each event a number between 0 and 1, with P(Ω)=1P(\Omega)=1P(Ω)=1, and probabilities of events that cannot happen together add up.

P(A∪B)=P(A)+P(B)−P(A∩B),P(Ac)=1−P(A)P(A\cup B)=P(A)+P(B)-P(A\cap B),\qquad P(A^{c})=1-P(A)P(A∪B)=P(A)+P(B)−P(A∩B),P(Ac)=1−P(A)(1)

Two events are mutually exclusive when A∩BA\cap BA∩B is empty, and independent when P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)P(A∩B)=P(A)P(B): knowing one tells you nothing about the other. These are different ideas: mutually exclusive events with positive probability are strongly dependent, because one happening rules the other out. For equally likely outcomes, P(A)P(A)P(A) is the number of outcomes in AAA divided by the total, so counting is the tool.

Counting and the complementTwo dice sum to 7 in 6 of 36 equally likely ways, so the probability is 6/36=1/66/36=1/66/36=1/6. Three independent trades each win with probability 0.6. “” is awkward to add up, but its complement “” is a single product: 0.43=0.0640.4^3=0.0640.43=0.064, so P(at least one)=1−0.064=0.936P(\text{at least one})=1-0.064=0.936P(at least one)=1−0.064=0.936.
  • Sample space: every possible outcome. Event: a set of outcomes.
  • Mutually exclusive: cannot happen together. Independent: P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)P(A∩B)=P(A)P(B).
  • Complement: P(Ac)=1−P(A)P(A^{c})=1-P(A)P(Ac)=1−P(A); use it for “”.
Exclusive does not mean independent“” and “” are mutually exclusive, so knowing it closed up tells you for sure it did not close down: they are as dependent as events can be. Independence needs the product rule, and is what you assume (often wrongly) when you multiply probabilities of losses across positions.

Lab · what's graded

  • Counting outcomes15 pts
  • The addition rule15 pts
  • Mutually exclusive events15 pts
  • Complement and independence15 pts
  • The vocabulary15 pts
  • At least one winner15 pts
  • Why use the complement10 pts
0/ 7 answered
Tasks
1Numeric answer15 ptsNot answered

Two fair dice are rolled. What is the probability that they sum to 7? Give it to four decimals.

P=P =P=accepted: ± 0.0005
2Numeric answer15 ptsNot answered

P(A)=0.5P(A)=0.5P(A)=0.5, P(B)=0.4P(B)=0.4P(B)=0.4 and P(A∩B)=0.2P(A\cap B)=0.2P(A∩B)=0.2. Find P(A∪B)P(A\cup B)P(A∪B).

P(A∪B)=P(A\cup B) =P(A∪B)=accepted: ± 0.001
3Choice15 ptsNot answered

Which pair of events is mutually exclusive for one stock on one day?

4Fill the blanks15 ptsNot answered

Let P(A)=0.35P(A)=0.35P(A)=0.35 and P(B)=0.4P(B)=0.4P(B)=0.4, with AAA and BBB independent. Fill in the blanks.

P(Ac)=,P(A∩B)=\displaystyle P(A^{c})=\htmlData{blank=p}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}},\qquad P(A\cap B)=\htmlData{blank=q}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}P(Ac)=,P(A∩B)=
5Match15 ptsNot answered

Match each term to its meaning.

Pick the partner of each item on the left.

Sample space
Event
Complement
Union
6Numeric answer15 ptsNot answered

Three independent trades each win with probability 0.6. What is the probability that at least one wins?

P=P =P=accepted: ± 0.001
7Explain in words10 ptsNot answered

In two sentences, say how you compute the probability of at least one winner among several independent trades and why that route is easier.

Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

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