CM2 April 2022 Q4

[Constantinos Spyrou]

Hello

I kindly ask for an explanation in question 4 on the past paper of April 2022 . When trying to apply Ito's lemma why is the partial derivative with respect to time:
-a'(T-t)f(t,Xt)?

If 'a' is a differentiable function with respect to t why is the product rule not being applied here?

 

[Steve Hales]

The function f is given as: exp(a(T-t)-x). The partial derivative with respect to t requires the chain rule because we're taking the exponential of a function.
Let me know how you would implement the product rule.

 

[Constantinos Spyrou]

Thanks for your reply
f(Xt,t)=exp(a(T-t)-Xt)
Hence,
partial(f)/partial(t)=a'(T-t)f(Xt,t) -af(Xt,t)


Clearly I miss something here, since I don't understand why the chain rule kicks in.

 

[Steve Hales]

Here's the chain rule (function-of-a-function rule): if f(t) = g(h(t)), then df/dt = dg/dh * dh/dt.
In the current question we have h(t) = a(T-t) - x, and g(h(t)) = exp(h(t)). Then:
partial(f)/partial(t) = dg/dh * dh/dt = f(t) * (-a'(T-t))

 

[Sunil Chaudhary]

Here's the chain rule (function-of-a-function rule): if f(t) = g(h(t)), then df/dt = dg/dh * dh/dt.
In the current question we have h(t) = a(T-t) - x, and g(h(t)) = exp(h(t)). Then:
partial(f)/partial(t) = dg/dh * dh/dt = f(t) * (-a'(T-t))

Hi Steve,

Here, I think I am also missing some thing as for me dh(t)/dt is coming as a'(T-t) + a(-1).
Can you please point out what I am missing here.

Thanks

 

[Steve Hales]

Hi.
You need the apply the chain rule to a(T-t), but it looks as though you've gone for the product rule instead. The question says that a is a differentiable function. Then, using the chain rule, we have:
dh(t)/dt = dh(t)/da * da/dt = a'(T-t) * (-1) = -a'(T-t)

 

[Sunil Chaudhary]

So, a(T-t) is a differentiable function of (T-t).

Further, will it make more sense if I write as below:
dh(t)/dt = d[a(T-t) - x]/d[a(T-t)] * d[a(T-t)]/d[(T-t)] * d[T-t]/dt = 1 * a'(T-t) * (-1) = -a'(T-t)