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I'm trying to solve a problem similar to Stochastic Integral. I have to evaluate $$ \mathbb{Var}\left(\int_{0}^t ((B_s)^2 + s) \mathrm{d}B_s \right)$$

I have split the problem in two parts:

1) $ \mathbb{Var}\left(\int_{0}^t (B_s)^2 \mathrm{d}B_s \right) = \mathbb{E}(\left(\int_{0}^t (B_s)^2 \mathrm{d}B_s \right)^2) - \mathbb{E}^2\left(\int_{0}^t (B_s)^2 \mathrm{d}B_s \right) = \left(\int_{0}^t \mathbb{E}(B_s)^4 \mathrm{d}s \right) - 0 = \left(\int_{0}^t 3t^2 \mathrm{d}s \right)= t^3$

Is this right?

2) $ \mathbb{Var}\left(\int_{0}^t s \mathrm{d}B_s \right) = \mathbb{E}(\left(\int_{0}^t s \mathrm{d}B_s \right)^2) - \mathbb{E}^2\left(\int_{0}^t s \mathrm{d}B_s \right) =\left(\int_{0}^t \mathbb{E}(s^2) \mathrm{d}s \right) - 0 = \left(\int_{0}^t s^2 \mathrm{d}s \right) = t^3$

And is this right?

So $$ \mathbb{Var}\left(\int_{0}^t ((B_s)^2 + s) \mathrm{d}B_s \right) = t^3 + t^3 = 2t^3$$

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Brownians
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    Do you believe that $\int_{0}^t s^2 \mathrm{d}s= t^3$ and that the variance of a sum is the sum of the variances, always? I am asking because you used both in your post... // OK, so your reaction is to mutilate your post, so now we do not know what you think... – Did Nov 26 '13 at 12:22

1 Answers1

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Hint Recall that for any progressively measurable function $f$

$$M_t := \int_0^t f(s) \, dB_s, \qquad t \leq T$$

is a martingale whenever

$$\mathbb{E} \left( \int_0^T |f(s)|^2 \, ds \right)<\infty$$

In particular, $$\mathbb{E}(M_t) = 0. \tag{1}$$ Moreover, Itô's isometry states

$$\mathbb{E} \left[ \left( \int_0^t f(s) \, dB_s \right)^2 \right] = \mathbb{E} \left( \int_0^t |f(s)|^2 \, ds \right) \tag{2}$$

Using $(1)$, $(2)$ for the function $f(s) := s+B_s^2$ allows you to compute the variance you are looking for.

saz
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  • I think the "error" that you point out was only a notation error. I have edited and completed my calculation but I'm not sure if it is correct, can you please check it out? – Brownians Nov 26 '13 at 11:12
  • @Brownians As far as I can see, you applied Itô's formula, but I fail to see why this should simplify the calculations. In any case, you have to use $(1)$ and $(2)$. – saz Nov 26 '13 at 11:58
  • I was just trying different things. I think I have the solution now. – Brownians Nov 26 '13 at 12:14
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    @Brownians 2nd calculation: $1/3$ is missing. And as Did already pointed out: In general, the variance of a sum is not the sum of the variances. – saz Nov 26 '13 at 13:10