By Eric Chin

Difficulties and ideas in Mathematical Finance: Stochastic Calculus (The Wiley Finance sequence) Mathematical finance calls for using complex mathematical recommendations drawn from the idea of likelihood, stochastic techniques and stochastic differential equations. those components are ordinarily brought and constructed at an summary point, making it frustrating whilst employing those innovations to sensible matters in finance. difficulties and strategies in Mathematical Finance quantity I: Stochastic Calculus is the 1st of a four-volume set of books targeting difficulties and ideas in mathematical finance. This quantity introduces the reader to the fundamental stochastic calculus options required for the research of this crucial topic, offering loads of labored examples which permit the reader to construct the required beginning for simpler oriented difficulties within the later volumes. via this program and through operating in the course of the a number of examples, the reader will appropriately comprehend and get pleasure from the basics that underpin mathematical finance. Written frequently for college students, practitioners and people considering educating during this box of research, Stochastic Calculus presents a important reference booklet to counterpoint one’s additional realizing of mathematical finance.

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**Problems and Solutions in Mathematical Finance Stochastic Calculus**

Difficulties and suggestions in Mathematical Finance: Stochastic Calculus (The Wiley Finance sequence) Mathematical finance calls for using complicated mathematical innovations drawn from the speculation of likelihood, stochastic procedures and stochastic differential equations. those components are quite often brought and constructed at an summary point, making it difficult whilst making use of those concepts to functional matters in finance.

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**Additional resources for Problems and Solutions in Mathematical Finance Stochastic Calculus**

**Example text**

2 Discrete and Continuous Random Variables 13 By differentiating MX (t) with respect to t twice we have M ′X (t) = npet (1 − p + pet )n−1 M ′′X (t) = npet (1 − p + pet )n−1 + n(n − 1)p2 e2t (1 − p + pet )n−2 and hence ????(X) = M ′X (0) = np Var(X) = ????(X 2 ) − ????(X)2 = M ′′X (0) − M ′X (0)2 = np(1 − p). Given the sequence Xi ∼ Bernoulli(p), i = 1, 2, . . , n are independent and identically distributed, each having expectation ???? = p and variance ???? 2 = p(1 − p), then as n → ∞, from the central limit theorem ∑n i=1 Xi − n???? D −−→ ????(0, 1) √ ???? n or D X − np −−→ ????(0, 1).

Prove that if n ∑ Y= Xi i=1 then Y follows a gamma distribution, Y ∼ Gamma (n, ????) with the following probability density function: (????y)n−1 −????y fY (y) = ????e , y ≥ 0. (n − 1)! Show also that ????(Y) = n n and Var(Y) = 2 . ???? ???? Solution: From the definition of the cumulative distribution function of W = U + V we obtain f (u, ????) dud???? FW (????) = ℙ(W ≤ ????) = ℙ(U + V ≤ ????) = ∫ ∫ UV u+????≤???? where fUV (u, ????) is the joint probability density function of (U, V). Since U ⟂ ⟂ V therefore fUV (u, ????) = fU (u)fV (????) and hence FW (????) = fUV (u, ????) dud???? ∫ ∫ u+????≤???? = fU (u)fV (????) dud???? ∫ ∫ u+????≤???? ∞ = ∫−∞ { ????−u ∫−∞ } fV (????) d???? fU (u) du ∞ = ∫−∞ FV (???? − u)fU (u) du.

3. Poisson Distribution. A discrete Poisson distribution, Poisson(????) with parameter ???? > 0 has the following probability mass function: ℙ(X = k) = ????k −???? e , k! k = 0, 1, 2, . . Show that ????(X) = ???? and Var(X) = ????. For a random variable following a binomial distribution, Binomial(n, p), 0 ≤ p ≤ 1 show that as n → ∞ and with p = ????∕n, the binomial distribution tends to the Poisson distribution with parameter ????. Solution: From the definition of the moment generating function ( ) ∑ tx MX (t) = ???? etX = e ℙ(X = x) x where t ∈ ℝ and substituting ℙ(X = x) = MX (t) = ∞ ∑ etx x=0 ????x −???? e we have x!