Prove that Y(t) = W(t)^2 - t is a martingale using the Ito lemma

Ito's Lemma:


If you have a function f that depends on time and another variable x, and if you differentiate it with respect to both time and x, you get the change in f. For our purposes, x is going to be our Brownian motion W(t).


We're interested in the function f(t, W(t)) = W(t)^2 - t.


For a function f(t, X(t)), the differential df using Ito's lemma is:


df = (∂f/∂t) dt + (∂f/∂X) dX + 0.5 (∂^2f/∂X^2) (dX)^2




- ∂f/∂t is the partial derivative of f with respect to t.

- ∂f/∂X is the first partial derivative of f with respect to X(t).

- ∂^2f/∂X^2 is the second partial derivative of f with respect to X(t).


To determine if Y(t) is a martingale, we'll find its differential and check its properties.


1. First Partial Derivative with respect to t:

The function Y(t) has a term -t, which directly depends on t. So, its derivative with respect to t is -1.


2. First Partial Derivative with respect to W(t):

Differentiating W(t)^2 with respect to W(t) gives 2 * W(t).


3. Second Partial Derivative with respect to W(t): 


Differentiating the result from the previous step, 2 * W(t), again with respect to W(t) gives 2.


Now, plug these values into Ito's Lemma:


dY(t) = (-1)dt + 2W(t)dW(t) + 0.5 * 2d(W(t))^2


dY(t) = -dt + 2W(t)dW(t) + d(W(t))^2



dY(t) = -dt + 2W(t)dW(t) + dt= 2W(t)dW(t)


(given that d(W(t))^2 equals dt)


For Y(t) to be a martingale, its expected change should be zero. From the equation above, the expectation of 2W(t)d(W(t)) is zero, as d(W(t)) has a mean of 0. 


Thus, we have a process Y(t) whose change consists solely of a diffusion term with zero expectation, which means that (Y(t)) does not drift up or down over time. This is precisely the property of a martingale.


This means Y(t) = W(t)^2 - t   is a martingale.


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