# A Course in Financial Calculus by Alison Etheridge

By Alison Etheridge

This article is designed for first classes in monetary calculus geared toward scholars with a great heritage in arithmetic. Key innovations resembling martingales and alter of degree are brought within the discrete time framework, permitting an obtainable account of Brownian movement and stochastic calculus. The Black-Scholes pricing formulation is first derived within the easiest monetary context. next chapters are dedicated to expanding the monetary sophistication of the types and tools. the ultimate bankruptcy introduces extra complicated themes together with inventory expense versions with jumps, and stochastic volatility. loads of routines and examples illustrate how the equipment and ideas will be utilized to lifelike monetary questions.

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**Extra info for A Course in Financial Calculus**

**Example text**

6 Suppose that {X n }n≥0 is a (P, {Fn }n≥0 )-submartingale. Then there is a previsible, non-decreasing process {An }n≥0 such that {X n − An }n≥0 is a (P, {Fn }n≥0 )martingale. If we insist that A0 = 0, then {An }n≥0 is unique. 2 Suppose that {X n }n≥0 is a (P, {Fn }n≥0 )-supermartingale. Then there is a previsible, non-decreasing process {An }n≥1 such that {X n + An }n≥0 is a (P, {Fn }n≥0 )martingale. If we insist that A0 = 0, then {An }n≥0 is unique. 1 Proof: The proofs of the two parts are essentially identical, so we restrict our attention to 1.

Suppose that there were another predictable process {Bn }n≥0 with the same property. Then {X n − An }n≥0 and {X n − Bn }n≥0 are both martingales and, therefore, so is the difference between them, {An − Bn }n≥0 . On the other hand {An − Bn }n≥0 is predictable and predictable martingales are constant (see Exercise 12). Since A0 = 0 = B0 , the proof is complete. ✷ American options and supermartingales Let’s see what these concepts correspond to in a ﬁnancial example. 2 and let Q be the probability measure on the tree under which the discounted stock price { S˜n }0≤n≤N is a martingale.

We must check that {X n − An }n≥0 is a martingale. First we check that E [|X n − An |] < ∞ for all n. E [|X n − An |] ≤ E [|X n |] + E [An ] n = E [|X n |] + E A0 + A j − A j−1 j=1 n = E [|X n |] + E E X j − X j−1 F j−1 (by deﬁnition of A j ) j=1 n ≤ E [|X n |] + E E X j + X j−1 F j−1 j=1 n = E [|X n |] + E X j + X j−1 (tower property), j=1 and evidently this ﬁnal expression is ﬁnite since by assumption E X j all j. Now we check the martingale property, < ∞ for E X n+1 − An+1 | Fn = E X n+1 − E X n+1 − X n | Fn − An Fn = E X n+1 − X n+1 + X n − An | Fn = X n − An .