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Probability Review for Financial Engineers
Part 2
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Conditional Probability
The conditional probability that E occurs given that F has occurred is denoted by π(πΈ|πΉ) If P(F) > 0 then π πΈ πΉ = π(πΈπΉ) π(πΉ)
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Example β 2 dice 2 dice are rolled - a red dice and a green dice, What is the probability distribution for the total? 2 - 1/ * (1/36) β¦ 7 β 6 * (1/36) 11 β 2 (1/36) 12 β 1/36
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2 Dice example continued
What is the expected value? 7 Ex) What is the probability distribution function for the total given that the Green dice was a 3, that is P(T|G=3) 4 β 1/6 5 β 1/6 6 β 1/6 7 β 1/6 8 β 1/6 9 β 1/6
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Example) Playing Cards
Selecting a card a standard 52 playing card deck What is the probability of getting an ace? 4/52 = 1/13 What is the probability of getting a ace given that someone already removed a jack from the deck? 4/51 the removal of a jack means that a non-ace has been removed from the deck What is the probability of getting an ace given that someone already removed a spade from the deck? 1/13 the removed cards suit is independent of the rank question.
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Joint Cumulative Distributions
F(a,b) = P(Xβ€a, Yβ€b) The distribution of X can be obtained from the joint distribution of X and Y as follows πΉ π =π π<π =π π<π|π< β =π( lim πββ π<π|π< π ) = lim πββ π π<π|π< π = lim πββ πΉ(π,π) =πΉ(π,β)
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Example β Time between arrivals
A market buy order and a market sell order arrive uniformly distributed between 1 and 2pm. Each person puts a 10 minute time limit on each order. What is the probability that the trade will not be executed because of a timeout? This would be the P(B +10 < S) + P(S+10 < B) = 2 P(B +10 < S) =2 π΅+10<π π π,π ππ ππ =2 π΅+10<π π π΅ (π) π π (π )ππ ππ = π β ππ ππ = π β10 ππ = 25 36
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Expected Values of Joint Densities
Suppose f(x,y) is a joint distribution πΈ[π π β π ] = ββ β ββ β π π₯ β π₯ π(π₯,π¦)ππ₯ ππ¦ = ββ β ββ β π π₯ β π₯ π π (π₯) π π (π¦)ππ₯ ππ¦ = ββ β β π₯ π π (π¦)ππ¦ ββ β π π₯ π π (π₯)ππ₯ =πΈ[β π ]πΈ[π π ]
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Covariance of 2 Random Variables
πΆππ£ π,π =πΈ (πβπΈ π β πβπΈ π ] =πΈ ππ βπΈ π πβππΈ π +πΈ π πΈ[π] =πΈ ππ βπΈ π πΈ[π]βπΈ[π]πΈ π +πΈ π πΈ[π] =πΈ ππ β πΈ π πΈ[π] Note that is X and Y are independent, then the covariance = 0
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Variance of sum of random variables
πππ π+π = πΈ[ π+πβπΈ π+π 2 ] = πΈ[ π+πβπΈπβπΈπ 2 ] = πΈ[ πβπΈπ+πβπΈπ 2 ] = πΈ πβπΈπ 2 + πβπΈπ 2 +2 πβπΈπ πβπΈπ = πΈ πβπΈπ 2 ]+ πΈ[ πβπΈπ 2 +2πΈ[ πβπΈπ πβπΈπ ] πππ π+π = πππ π +πππ π +2πΆππ£(π,π)
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Correlation of 2 random variables
As long as Var(X) and Var(Y) are both positive, the correlation of X and Y is denotes as π π,π = πΆππ£(π,π) πππ π πππ(π) It can be shown that β1 β€ π π,π β€1 The correlation coefficient is a measure of the degree of linearity between X and Y π π,π =0 means very little linearity π π,π ππππ+1 means X and Y increase and decrease together π π,π ππππβ1 means X and Y increase and decrease inversely
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Central Limit Theorem Loosely put, the sum of a large number of independent random variables has a normal distribution. Let π 1 , π 2 β¦ be a sequence of independent and identically distributed random variables each having mean π and variance π 2 Then the distribution of π 1 +β¦+ π π βnπ π π Tends to a standard normal as nο β, that is π π 1 +β¦+ π π βnπ π π β€π β 1 2π ββ π π β π₯ 2 /2 ππ₯
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