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Normal Random Variables and Probability An Undergraduate Introduction to Financial Mathematics J. Robert Buchanan 2010 J. Robert Buchanan Normal Random Variables and Probability Discrete vs. Continuous Random Variables Think about the probability of selecting X from the interval [0, 1] when X ∈ {0, 1} J. Robert Buchanan Normal Random Variables and Probability Discrete vs. Continuous Random Variables Think about the probability of selecting X from the interval [0, 1] when X ∈ {0, 1} X ∈ {k/10 : k = 0, 1, . . . , 10} J. Robert Buchanan Normal Random Variables and Probability Discrete vs. Continuous Random Variables Think about the probability of selecting X from the interval [0, 1] when X ∈ {0, 1} X ∈ {k/10 : k = 0, 1, . . . , 10} X ∈ {k/n : k = 0, 1, . . . , n} J. Robert Buchanan Normal Random Variables and Probability Discrete vs. Continuous Random Variables Think about the probability of selecting X from the interval [0, 1] when X ∈ {0, 1} X ∈ {k/10 : k = 0, 1, . . . , 10} X ∈ {k/n : k = 0, 1, . . . , n} Question: what happens to the last probability as n → ∞? J. Robert Buchanan Normal Random Variables and Probability Continuous Random Variables Definition A random variable X has a continuous distribution (or probability distribution function or probability density function) if there exists a non-negative function f : R → R such that for an interval [a, b] the P (a ≤ X ≤ b) = Z b f (x) dx. a The function f must, in addition to satisfying f (x) ≥ 0, have the following property, Z ∞ f (x) dx = 1. −∞ J. Robert Buchanan Normal Random Variables and Probability Area Under the PDF fHxL x a J. Robert Buchanan b Normal Random Variables and Probability Uniformly Distributed Continuous Random Variables Definition A continuous random variable X is uniformly distributed in the interval [a, b] (with b > a) if the probability that X belongs to any subinterval of [a, b] is equal to the length of the subinterval divided by b − a. J. Robert Buchanan Normal Random Variables and Probability Uniformly Distributed Continuous Random Variables Definition A continuous random variable X is uniformly distributed in the interval [a, b] (with b > a) if the probability that X belongs to any subinterval of [a, b] is equal to the length of the subinterval divided by b − a. Question: Assuming the PDF vanishes outside of [a, b] and is constant on [a, b], what is the PDF? J. Robert Buchanan Normal Random Variables and Probability Example Example Random variable X is continuously uniformly randomly distributed in the interval [−5, 5]. Find the probability that −1 ≤ X ≤ 2. J. Robert Buchanan Normal Random Variables and Probability Example Example Random variable X is continuously uniformly randomly distributed in the interval [−5, 5]. Find the probability that −1 ≤ X ≤ 2. P (−1 ≤ X ≤ 2) = J. Robert Buchanan 3 2 − (−1) = 5 − (−5) 10 Normal Random Variables and Probability Expected Value Definition The expected value or mean of a continuous random variable X with probability density function f (x) is Z ∞ x f (x) dx. E [X ] = −∞ J. Robert Buchanan Normal Random Variables and Probability Example Example Find the expected value of X if X is a continuously uniformly distributed random variable on the interval [−10, 80]. J. Robert Buchanan Normal Random Variables and Probability Example Example Find the expected value of X if X is a continuously uniformly distributed random variable on the interval [−10, 80]. E [X ] = Z ∞ x f (x) dx −∞ Z 80 x dx −10 90 80 x 2 = 180 −10 6400 100 = − 180 180 = 35 = J. Robert Buchanan Normal Random Variables and Probability Expected Value of a Function Definition The expected value of a function g of a continuously distributed random variable X which has probability distribution function f is defined as Z ∞ g(x)f (x) dx, E [g(X )] = −∞ provided the improper integral converges absolutely, i.e., E [g(X )] is defined if and only if Z ∞ |g(x)|f (x) dx < ∞. −∞ J. Robert Buchanan Normal Random Variables and Probability Example Example Find the expected value of X 2 if X is continuously distributed on [0, ∞) with probability density function f (x) = e−x . J. Robert Buchanan Normal Random Variables and Probability Example Example Find the expected value of X 2 if X is continuously distributed on [0, ∞) with probability density function f (x) = e−x . h E X 2 i = = = = Z ∞ x 2 e−x dx 0 lim Z M M→∞ 0 lim M→∞ lim M→∞ = 2 h h x 2 e−x dx iM −(x 2 + 2x + 2)e−x 0 2 2 − (M + 2M + 2)e J. Robert Buchanan −M i Normal Random Variables and Probability Joint and Marginal Distributions Definition A joint probability distribution for a pair of random variables, X and Y , is a non-negative function f (x, y) for which Z ∞Z ∞ f (x, y) dx dy = 1. −∞ −∞ J. Robert Buchanan Normal Random Variables and Probability Joint and Marginal Distributions Definition A joint probability distribution for a pair of random variables, X and Y , is a non-negative function f (x, y) for which Z ∞Z ∞ f (x, y) dx dy = 1. −∞ −∞ Definition If X and Y are continuous random variables with joint distribution f (x, y) then the marginal distribution for X is defined as the function Z ∞ f (x, y) dy. fX (x) = −∞ J. Robert Buchanan Normal Random Variables and Probability Joint and Marginal Distributions Definition A joint probability distribution for a pair of random variables, X and Y , is a non-negative function f (x, y) for which Z ∞Z ∞ f (x, y) dx dy = 1. −∞ −∞ Definition If X and Y are continuous random variables with joint distribution f (x, y) then the marginal distribution for X is defined as the function Z ∞ f (x, y) dy. fX (x) = −∞ Remark: a similar definition may be stated for the marginal distribution for Y . J. Robert Buchanan Normal Random Variables and Probability Independence of Jointly Distributed RVs Definition Two continuous random variables are independent and only if the joint probability distribution function factors into the product of the marginal distributions of X and Y . In other words X and Y are independent if and only if f (x, y) = fX (x)fY (y) for all real numbers x and y. J. Robert Buchanan Normal Random Variables and Probability Example (1 of 2) Example Consider the jointly distributed random variables (X , Y ) ∈ [0, ∞) × [−2, 2] whose distribution is the function f (x, y) = 4e1x . Find the mean of X + Y . J. Robert Buchanan Normal Random Variables and Probability Example (2 of 2) E [X + Y ] = = = = Z 0 Z ∞Z 2 (x + y) −2 ∞ Z0 ∞ Z0 ∞ 1 4ex dy dx Z 1 −x 2 e (x + y) dy dx 4 −2 1 −x e (4x) dx 4 xe−x dx 0 = = lim Z M→∞ 0 M xe−x dx lim (1 − Me−M − e−M ) M→∞ = 1 J. Robert Buchanan Normal Random Variables and Probability Properties of the Expected Values Theorem If X1 , X2 , . . . , Xk are continuous random variables with joint probability distribution f (x1 , x2 , . . . , xk ) then E [X1 + X2 + · · · + Xk ] = E [X1 ] + E [X2 ] + · · · + E [Xk ]. J. Robert Buchanan Normal Random Variables and Probability Properties of the Expected Values Theorem If X1 , X2 , . . . , Xk are continuous random variables with joint probability distribution f (x1 , x2 , . . . , xk ) then E [X1 + X2 + · · · + Xk ] = E [X1 ] + E [X2 ] + · · · + E [Xk ]. Theorem Let X1 , X2 , . . . , Xk be pairwise independent random variables with joint distribution f (x1 , x2 , . . . , xk ), then E [X1 X2 · · · Xk ] = E [X1 ] E [X2 ] · · · E [Xk ] . J. Robert Buchanan Normal Random Variables and Probability Variance and Standard Deviation Definition If X is a continuously distributed random variable with probability density function f (x), the variance of X is defined as i Z ∞ h (x − µ)2 f (x) dx, Var (X ) = E (X − µ)2 = −∞ where µp = E [X ]. The standard deviation of X is σ(X ) = Var (X ). J. Robert Buchanan Normal Random Variables and Probability Variance and Standard Deviation Definition If X is a continuously distributed random variable with probability density function f (x), the variance of X is defined as i Z ∞ h (x − µ)2 f (x) dx, Var (X ) = E (X − µ)2 = −∞ where µp = E [X ]. The standard deviation of X is σ(X ) = Var (X ). Theorem Let X be a random variable with probability distribution f and mean µ, then Var (X ) = E X 2 − µ2 . J. Robert Buchanan Normal Random Variables and Probability Example Example Suppose X is continuously distributed on [0, ∞) with probability density function f (x) = e−x . Find Var (X ). J. Robert Buchanan Normal Random Variables and Probability Example Example Suppose X is continuously distributed on [0, ∞) with probability density function f (x) = e−x . Find Var (X ). h i Var (X ) = E X 2 − (E [X ])2 Z Z ∞ 2 −x x e dx − = 0 = 2− Z ∞ 0 = 2 − (1)2 xe−x dx ∞ 0 2 xe −x dx 2 = 1 J. Robert Buchanan Normal Random Variables and Probability Properties of Variance Theorem Let X be a continuous random variable with probability distribution f (x) and let a, b ∈ R, then Var (aX + b) = a2 Var (X ) . J. Robert Buchanan Normal Random Variables and Probability Properties of Variance Theorem Let X be a continuous random variable with probability distribution f (x) and let a, b ∈ R, then Var (aX + b) = a2 Var (X ) . Theorem Let X1 , X2 , . . . , Xk be pairwise independent continuous random variables with joint probability distribution f (x1 , x2 , . . . , xk ), then Var (X1 + X2 + · · · + Xk ) = Var (X1 ) + Var (X2 ) + · · · + Var (Xk ) . J. Robert Buchanan Normal Random Variables and Probability Normal Random Variable Assumption: any characteristic of an object subject to a large number of independently acting forces typically takes on a normal distribution. J. Robert Buchanan Normal Random Variables and Probability Normal Random Variable Assumption: any characteristic of an object subject to a large number of independently acting forces typically takes on a normal distribution. We will develop the normal probability density function from the probability function for the binomial random variable. P (X = x) = n! p x (1 − p)n−x x!(n − x)! J. Robert Buchanan Normal Random Variables and Probability Overview of Derivation Thought Experiment: Imagine standing at the origin of the number line and at each tick of a clock taking a step to the left or the right. In the long run where will you stand? J. Robert Buchanan Normal Random Variables and Probability Overview of Derivation Thought Experiment: Imagine standing at the origin of the number line and at each tick of a clock taking a step to the left or the right. In the long run where will you stand? Assumptions: 1 n steps/ticks, 2 random walk takes place during time interval [0, t], which implies a “tick” lasts ∆t, 3 on each tick move a distance ∆x > 0, 4 n(∆x)2 = 2kt, 5 probability of moving left/right is 1/2, 6 all steps are independent. J. Robert Buchanan Normal Random Variables and Probability Take a Few Steps Suppose r out of n steps (0 ≤ r ≤ n) have been to the right. Question: Where are you? J. Robert Buchanan Normal Random Variables and Probability Take a Few Steps Suppose r out of n steps (0 ≤ r ≤ n) have been to the right. Question: Where are you? (r − (n − r ))∆x = (2r − n)∆x = m∆x Question: What is the probability of standing there? J. Robert Buchanan Normal Random Variables and Probability Take a Few Steps Suppose r out of n steps (0 ≤ r ≤ n) have been to the right. Question: Where are you? (r − (n − r ))∆x = (2r − n)∆x = m∆x Question: What is the probability of standing there? P (X = m∆x) = P (X = (2r − n)∆x) r n−r n 1 1 = r 2 2 n 1 n! = r !(n − r )! 2 n n! 12 1 = 1 2 (n + m) ! 2 (n − m) ! J. Robert Buchanan Normal Random Variables and Probability Bernoulli Steps Each step is a Bernoulli experiment with outcomes ∆x and −∆x. Questions: What is the expected value of a single step? What is the variance in the outcomes? J. Robert Buchanan Normal Random Variables and Probability Bernoulli Steps Each step is a Bernoulli experiment with outcomes ∆x and −∆x. Questions: What is the expected value of a single step? E [X ] = 0 What is the variance in the outcomes? J. Robert Buchanan Normal Random Variables and Probability Bernoulli Steps Each step is a Bernoulli experiment with outcomes ∆x and −∆x. Questions: What is the expected value of a single step? E [X ] = 0 What is the variance in the outcomes? Var (X ) = (∆x)2 J. Robert Buchanan Normal Random Variables and Probability The Sum of Bernoulli Steps Questions: after n steps, What is the expected value of where you stand? What is the variance in final position? J. Robert Buchanan Normal Random Variables and Probability The Sum of Bernoulli Steps Questions: after n steps, What is the expected value of where you stand? " n # X X = nE [X ] = 0 E i=1 What is the variance in final position? J. Robert Buchanan Normal Random Variables and Probability The Sum of Bernoulli Steps Questions: after n steps, What is the expected value of where you stand? " n # X X = nE [X ] = 0 E i=1 What is the variance in final position? ! n X X = nVar (X ) = n(∆x)2 Var i=1 J. Robert Buchanan Normal Random Variables and Probability Stirling’s Formula n! ≈ √ 2πe−n nn+1/2 Replace all the factorials with Stirling’s Formula. J. Robert Buchanan Normal Random Variables and Probability Stirling’s Formula n! ≈ √ 2πe−n nn+1/2 Replace all the factorials with Stirling’s Formula. √ n 2πe−n nn+1/2 12 √ (n+m+1)/2 √ (n−m+ 2πe−(n+m)/2 12 (n + m) 2πe−(n−m)/2 12 (n − m) −(n+1)/2 m −m/2 m m/2 2 m2 1+ 1− = √ 1− 2 n n n 2nπ J. Robert Buchanan Normal Random Variables and Probability Further Simplification Since m = x/∆x and n = t/∆t, −(n+1)/2 m −m/2 m m/2 m2 √ 1+ 1− 1− 2 n n n 2nπ √ x x !− 1+t/∆t 2 x∆t − 2∆x x∆t 2 x∆t 2∆x 2 ∆t 1+ 1− 1− = √ t∆x t∆x t∆x 2πt x x " #− kt 2 − 12 x ∆x 2 (∆x ) ∆x x ∆x − 2∆x x ∆x 2∆x 1− = √ 1+ 1− , 2kt 2kt 2kt kπt 2 since (∆x)2 = 2k∆t. J. Robert Buchanan Normal Random Variables and Probability Passing to the Limit As we take the limit with ∆x → 0, the probability of standing at exactly one, specific location becomes 0. Instead we must change our thinking and ask for P ((m − 1)∆x < X < (m + 1)∆x) ≈ 2(∆x)f (x, t). J. Robert Buchanan Normal Random Variables and Probability Passing to the Limit As we take the limit with ∆x → 0, the probability of standing at exactly one, specific location becomes 0. Instead we must change our thinking and ask for P ((m − 1)∆x < X < (m + 1)∆x) ≈ 2(∆x)f (x, t). f (x, t) = = = −x x " #− k 1 x ∆x 2 (∆ x ∆x 2∆x x ∆x 2∆x √ 1− 1− lim 1 + 2kt 2kt 2kt 2 kπt ∆x→0 −kt 2 1 x − x2 − x x2 − x √ e 2kt e 4k 2 t 2 e 2kt 2 kπt x2 1 √ e− 4kt 2 kπt J. Robert Buchanan Normal Random Variables and Probability Is f (x, t) a PDF? Suppose Z ∞ −∞ S2 = = = = x2 1 √ e− 4kt dx = S, then 2 kπt Z ∞ Z ∞ y2 x2 1 1 √ √ e− 4kt dx e− 4kt dy −∞ 2 kπt −∞ 2 kπt Z ∞Z ∞ 1 2 2 e−(x +y )/4kt dx dy 4kπt −∞ −∞ Z 2π Z ∞ 2 1 re−r /4kt dr d θ 4kπt 0 0 1 J. Robert Buchanan Normal Random Variables and Probability Surface Plot The graph of the PDF resembles: z t x J. Robert Buchanan Normal Random Variables and Probability The Bell Curve For a fixed value of t, the graph of the PDF resembles: y x J. Robert Buchanan Normal Random Variables and Probability Expected Value and Variance If X is a continuously distributed random variable with PDF: x2 1 f (x, t) = √ e− 4kt 2 kπt then J. Robert Buchanan Normal Random Variables and Probability Expected Value and Variance If X is a continuously distributed random variable with PDF: x2 1 f (x, t) = √ e− 4kt 2 kπt then E [X ] = Z ∞ −∞ x2 x √ e− 4kt dx = 0 2 kπt and J. Robert Buchanan Normal Random Variables and Probability Expected Value and Variance If X is a continuously distributed random variable with PDF: x2 1 f (x, t) = √ e− 4kt 2 kπt then E [X ] = Z ∞ −∞ and Var (X ) = Z ∞ −∞ x2 x √ e− 4kt dx = 0 2 kπt x2 x2 √ e− 4kt dx − (E [X ])2 = 2kt 2 kπt σ2 and thus 2kt = and we express the PDF for a normally distributed random variable with mean µ and variance σ 2 as 2 1 − (x −µ) f (x) = √ e 2σ2 . σ 2π J. Robert Buchanan Normal Random Variables and Probability Standard Normal Distribution x2 1 When µ = 0 and σ = 1 the PDF f (x) = √ e− 2 is called the 2π standard normal distribution. J. Robert Buchanan Normal Random Variables and Probability Standard Normal Distribution x2 1 When µ = 0 and σ = 1 the PDF f (x) = √ e− 2 is called the 2π standard normal distribution. The cumulative distribution function φ(x) is defined as Z x t2 1 √ e− 2 dt. φ(x) = P (X < x) = 2π −∞ J. Robert Buchanan Normal Random Variables and Probability Change of Variable Theorem If X is a normally distributed random variable with expected value µ and variance σ 2 , then Z = (X − µ)/σ is normally distributed with an expected value of zero and a variance of one. J. Robert Buchanan Normal Random Variables and Probability Central Limit Theorem (1 of 2) Suppose the random variables X1 , X2 , . . . , Xn 1 2 are pairwise independent but not necessarily identically distributed, have means µ1 , µ2 , . . . , µn and variances σ12 , σ22 , . . . , σn2 , and we define a new random variable Yn as Pn i=1 (Xi − µi ) . Yn = q Pn 2 σ i=1 i J. Robert Buchanan Normal Random Variables and Probability Central Limit Theorem (1 of 2) Suppose the random variables X1 , X2 , . . . , Xn 1 2 are pairwise independent but not necessarily identically distributed, have means µ1 , µ2 , . . . , µn and variances σ12 , σ22 , . . . , σn2 , and we define a new random variable Yn as Pn i=1 (Xi − µi ) . Yn = q Pn 2 σ i=1 i A Central Limit Theorem due to Liapounov implies that Yn has the standard normal distribution. J. Robert Buchanan Normal Random Variables and Probability Central Limit Theorem (2 of 2) Theorem Suppose that the infinite collection {Xi }∞ i=1 of random variables are pairwiseindependent and that for each i ∈ N we have E |Xi − µi |3 < ∞. If in addition, lim n→∞ Pn then for any x ∈ R 3 i=1 E |Xi − µi | Pn 2 3/2 i=1 σi =0 lim P (Yn ≤ x) = φ(x) n→∞ where random variable Yn is defined as above. J. Robert Buchanan Normal Random Variables and Probability Example (1 of 2) Example Suppose the annual snowfall in Millersville, PA is 14.6 inches with a standard deviation of 3.2 inches and is normally distributed. Snowfall amounts in different years are independent. What is the probability that the sum of the snowfall amounts in the next two years will exceed 30 inches? J. Robert Buchanan Normal Random Variables and Probability Example (2 of 2) Solution: If X represents the random variable standing for the snowfall received in Millersville, PA for one year then X + X is the random variable representing the snowfall of two years. The random variable X + X has mean µ = 2(14.6) = 29.2 inches and variance σ 2 = (3.2)2 + (3.2)2 . ! 30 − 29.2 P (X + X > 30) = P Z > p (3.2)2 + (3.2)2 = 1 − P (Z ≤ 0.176777) = 1 − φ(0.176777) = 0.429842 J. Robert Buchanan Normal Random Variables and Probability Lognormal Random Variables Definition A random variable X is a lognormal random variable with parameters µ and σ if ln X is a normally distributed random variable with mean µ and variance σ 2 . J. Robert Buchanan Normal Random Variables and Probability Lognormal Random Variables Definition A random variable X is a lognormal random variable with parameters µ and σ if ln X is a normally distributed random variable with mean µ and variance σ 2 . Remarks: The parameters µ is sometimes called the drift. The parameter σ is sometimes called the volatility. J. Robert Buchanan Normal Random Variables and Probability Lognormal PDF (1 of 2) Suppose X is lognormal, then Y = ln X is normal and P (X < x) = P (Y < ln x) Z ln x 1 2 2 = √ e−(t−µ) /2σ dt 2π −∞ If we let u = et and du = et dt, then Z x 1 1 −(ln u−µ)2 /2σ2 P (Y < ln x) = √ e du 2π −∞ u J. Robert Buchanan Normal Random Variables and Probability Lognormal PDF (2 of 2) f (x) = 1 2 2 √ e−(ln x−µ) /2σ (σ 2π)x y x J. Robert Buchanan Normal Random Variables and Probability Mean and Variance of a Lognormal RV Lemma If X is a lognormal random variable with parameters µ and σ then E [X ] = eµ+σ 2 /2 Var (X ) = e2µ+σ J. Robert Buchanan 2 2 eσ − 1 Normal Random Variables and Probability Derivation of Mean of a Lognormal RV Z ∞ 1 1 −(ln x−µ)2 /2σ2 √ e dx E [X ] = x x σ 2π 0 Z ∞ 1 2 2 √ = et e−(t−µ) /2σ dt σ 2π −∞ Z ∞ 1 2 2 2 µ+σ2 /2 √ e−(t−(µ+σ )) /2σ dt = e σ 2π −∞ = eµ+σ 2 /2 J. Robert Buchanan Normal Random Variables and Probability Derivation of Variance of a Lognormal RV h i Var (X ) = E X 2 − (E [X ])2 Z ∞ 2 1 2 2 1 −(ln x−µ)2 /2σ2 √ x = e dx − eµ+σ /2 x σ 2π 0 Z ∞ 1 2 2 √ e2t e−(t−µ) /2 dt − e2µ+σ = σ 2π −∞ Z ∞ 1 2 2 2(µ+σ2 ) √ = e e−(t−(µ+2σ)) /2 dt − e2µ+σ σ 2π −∞ 2 2 = e2µ+σ eσ − 1 J. Robert Buchanan Normal Random Variables and Probability Lognormal RVs and Security Prices Observation: Let S(0) denote the price of a security at some starting time arbitrarily chosen to be t = 0. For n ≥ 1, let S(n) denote the price of the security on day n. The random variable X (n) = S(n)/S(n − 1) for n ≥ 1 is lognormally distributed, i.e., ln X (n) = ln S(n) − ln S(n − 1) is normally distributed. J. Robert Buchanan Normal Random Variables and Probability Closing Prices of Sony (SNE) Stock Closing prices of Sony Corporation stock: 55 50 45 40 Oct Jan J. Robert Buchanan Apr Jul Normal Random Variables and Probability Lognormal Behavior of Sony (SNE) Stock Lognormal behavior of closing prices: 30 25 20 15 10 5 -0.04 -0.02 0 0.02 0.04 0.06 µ = 0.000416686 σ = 0.0160606 J. Robert Buchanan Normal Random Variables and Probability Example (1 of 2) Example What is the probability that the closing price of Sony Corporation stock will be higher today than yesterday? J. Robert Buchanan Normal Random Variables and Probability Example (1 of 2) Example What is the probability that the closing price of Sony Corporation stock will be higher today than yesterday? S(n) S(n) = P P > 1 > ln 1 ln S(n − 1) S(n − 1) | {z } {z } | lognormal normal = P (X > 0) 0 − 0.000416686 = P Z > 0.0160606 = 1 − P (Z ≤ −0.0259446) = 1 − φ(−0.0259446) = 0.510349 J. Robert Buchanan Normal Random Variables and Probability Example (2 of 2) Example What is the probability that tomorrow’s closing price will be higher than yesterday’s closing price? J. Robert Buchanan Normal Random Variables and Probability Example (2 of 2) Example What is the probability that tomorrow’s closing price will be higher than yesterday’s closing price? P S(n + 1) >1 S(n − 1) = = = = = S(n + 1) S(n) P >1 S(n) S(n − 1) S(n) S(n + 1) + ln >0 P ln S(n) S(n − 1) P (X + X > 0) 0 − 2(0.000416686) P Z >√ 0.01606062 + 0.01606062 1 − P (Z ≤ −0.0366912) = 1 − φ(−0.0366912) = 0.514634 J. Robert Buchanan Normal Random Variables and Probability Properties of Expected Value and Variance If an item is worth K but can only be sold for X , a rational investor would sell only if X ≥ K . The payoff of the sale can be expressed as X − K if X ≥ K , (X − K )+ = 0 if X < K . J. Robert Buchanan Normal Random Variables and Probability Payoff When X is Normal Corollary If X is normal random variable with mean µ and variance σ 2 and K is a constant, then µ−K σ −(µ−K )2 /2σ2 + + (µ − K )φ E (X − K ) = √ e , σ 2π Var (X − K )+ µ − 2K (µ − 2K )σ 2 2 2 2 √ e−(µ−2K ) /2σ + = (µ − 2K ) + σ φ σ 2π 2 σ −(µ−K )2 /2σ2 µ−K . − √ e + (µ − K )φ σ 2π J. Robert Buchanan Normal Random Variables and Probability Payoff When X is Lognormal Corollary If X is a lognormally distributed random variable with parameters µ and σ 2 and K > 0 is a constant then µ − ln K µ − ln K + µ+σ2 /2 E (X − K ) = e φ + σ − Kφ , σ σ Var (X − K )+ 2 = e2(µ+σ ) φ(w + 2σ) + K 2 φ(w ) where w = (µ − ln K )/σ. 2 − 2Keµ+σ /2 φ(w + σ) 2 2 − eµ+σ /2 φ(w + σ) − K φ(w ) J. Robert Buchanan Normal Random Variables and Probability

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