# Ito's formula for $B^2$ doesn't work with Monte Carlo

I am trying to "verify" Ito's formula for $B^2$ with simple Monte Carlo method.
According to my textbook, $B^2$ satisfies this SDE: $d(B^2) = dt + 2BdB$.

My simulation (notebook) results in following distribution for $B^2$ with $dt \in \{1/100,1/1000,1/10000\}$:

And distribution for solution of $dy = dt + 2BdB$:

Clearly they are different and it looks like I am making some very simple mistake, am I?

Here's my code in Julia (notebook):

# k instances of ito integral simulations of f from t=0. to t=1.
function ito_sim(f, n, k)
y   = zeros(k)
dt  = 1./n
sdt = sqrt(dt)
t = 0.
b = zeros(k)
for i in 1:n
db = randn(k) * sdt
b += db
t += dt
y += f(t, dt, b, db)
end
y
end;
plt[:hist](ito_sim((t,dt,b,db)-> db, n, 10000) .^ 2);
plt[:hist](ito_sim((t,dt,b,db)-> dt + 2 * b .* db, n, 10000));


PS I searched through stackoverflow and it seem to me that math.stachexchange is better suited for this question.

Looking at your code , your scheme is biased , and it is because of the way you calculate the Ito integral.

You built a function that calculate samples of an Ito process at time $1$. Let $g(t,B_t)$ an Ito process defined as

$$dg(t,B_t)=a(t,B_t)dt+\sigma(t,B_t)dB_t$$ where $B_t$ is a Brownian motion.

You say that given a subdivision $\{t_i\}$, defined as $t_i=\frac{i}{n}$

$$g(1,B_1)=\sum_{k=1}^{n}{[g(t_k,B_{t_k})-g(t_{k-1},B_{t_{k-1}})]}\approx\sum_{k=1}^{n}{[a(t_k,B_{t_{k}})(t_k-t_{k-1})+\sigma(t_k,B_{t_{k}})(B_{t_k}-B_{t_{k-1}})]}$$

Then we define a function $$f(t,dt,b,db)=[a(t,b)dt+\sigma(t,b)db]$$

$$g(1,B_1)\approx\sum_{k=1}^{n}{f(t_k,t_k-t_{k-1},B_{t_{k}},B_{t_{k}}-B_{t_{k-1}})}$$

The above formula can be a good approximation if $g(0,B_0)=0$, which is the case here. When you apply your code to the process $$dg(t,B_t)=dt+2B_tdB_t$$

Your final result will look like

$$X=\sum_{k=1}^{n}{[\frac{1}{n}+2B_{t_{k}}(B_{t_k}-B_{t_{k-1}})]}$$ and its expectation is $3$, while you expect $1$.

The reason is that you use this approximation in your code $$\int_{t_k}^{t_{k+1}}{B_tdB_t}\approx B_{t_{k+1}}(B_{t_{k+1}}-B_{t_{k}})$$

$$\int_{t_k}^{t_{k+1}}{B_tdB_t}\approx B_{t_{k}}(B_{t_{k+1}}-B_{t_{k}})$$
the bias is removed, and the expectation would have been $1$.
• I think you are right that there's a bias involved. I remember something in one of the books. The average of $B_k$ and $B_{k+1}$ doesn't work though. But using $B_k$ does! I am getting exactly the distribution I wanted: i.imgur.com/CeFr8Ao.png – mikea Dec 26 '16 at 3:05
• Exactly, corrected. When we choose $B_{t_k}$, we have the right result – Canardini Dec 26 '16 at 4:05