Quadratic Variation in SDEs. Let Xt be a stochastic process satisfying: X_t=X_0+int_0^t mu (s, omega)ds+int_0^t ν(s,omega)dB_s

Pellagra3d

Pellagra3d

Answered question

2022-10-27

Quadratic Variation in SDEs.
Let X t be a stochastic process satisfying:
X t = X 0 + 0 t μ ( s , ω ) d s + 0 t ν ( s , ω ) d B s
Shorthand: d X t = μ t d t + ν t d B t
where the last integral is a Brownian motion integral, and where μ ( t , ω ) , ν ( t , ω ) are F t adapted L 2 functions. (In class we so far only did the case where μ and ν are deterministic).
Let f(t,x) be a twice differentiable deterministic function.
The usual presentation of Ito's lemma is:
d f ( t , X t ) = ( f t + μ t f x + 1 2 2 f x 2 ν t 2 ) d t + f x ν t d B t
The professor offered us this shorthand:
d f ( t , X t ) = f t d t + f x d X t + 1 2 2 f 2 x ( d X t ) 2
He explained that the notation ( d X t ) 2 is to be interpreted as the taking the quadratic variation whenever the algebra would suggest you multiply the differentials. For example, d B t d B t = d B t , B t T = d ( T ) a.s. = d t (this last step has its own answer on s.e.) Why is the formal multiplication of (stochastic) differentials interpreted as the quadratic variation?

Answer & Explanation

Steinherrjm

Steinherrjm

Beginner2022-10-28Added 12 answers

Step 1
Well, to my modest opinion, we only have the slightly boring answer: Because it works. Via the ito isommetry one can show, that for any bounded, predictable f 1 , f 2 and semimartingales X, Y one has:
0 f s 1 d X s , 0 f s 2 d Y s t = 0 t f s 1 f s 2 d X , Y s
Step 2
which is basically the justification of doing it. Furthermore, one can even make it rigorous with some algebraic structures.

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