A Factor Model Approach to Derivative Pricing by James A. Primbs

By James A. Primbs

This e-book presents the quickest and easiest path to the vast majority of the implications and equations in by-product pricing, and provides the reader the instruments essential to expand those principles to new events that they could come across. It does so through targeting a unmarried underlying precept that's effortless to understand, after which it indicates that this precept is the most important to the vast majority of the implications in spinoff pricing. In that experience, it presents the "big photo" of spinoff pricing via targeting the underlying precept and never on mathematical technicalities. After interpreting this ebook, one is provided with the instruments had to expand the strategies to any new pricing state of affairs.

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Note: In this problem we converge to a Poisson random variable with parameter α since we took the time interval to be 1. If the time interval is t, we will converge to a Poisson random variable with parameter αt. 5 Poisson Processes Again. Consider the following Markov chain. Let the state space be the whole numbers 16 A FACTOR MODEL APPROACH TO DERIVATIVE PRICING x = 0, 1, 2, .... Consider the following transition probabilities over the time instant dt: P(x(t + dt) = n|x(t) = n) = 1 − αdt. P(x(t + dt) = n + 1|x(t) = n) = αdt.

5 Poisson jump at time s. 15). But this indicates that π never jumps since this equation holds for every t! Something must be wrong! What is wrong is that we have assumed that π is right continuous, and then when we add dt to the current time, we are implicitly taking the limit from the right. Hence, we are guaranteed never to capture the jump! This problem arises from our convention of assuming that Poisson processes are right continuous. If we had assumed left continuity, we would have avoided this specific issue.

Let’s approximate the integral above by the sum S(T, ∆t) = T ∆t −1 i=0 T 2 (z((i + 1)∆t) − z(i∆t)) ≈ dz 2 . 12) 0 Now, the claim is that as ∆t → 0, then the sum S(T, ∆t) → T . But note that S(T, ∆t) is a random variable since it involves z(t). Therefore, we are trying to show that the random variable S(T, ∆t) converges to the constant T . To do this, we will show that the mean of S(T, ∆t) is equal to T , and its variance approaches zero. This does the trick, since a random variable with zero variance must be a constant equal to its mean.

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