Change of measure method

Question

Let $f(t)=t^2+1$,

Using change of measure method calculate

$$\mathbb{E}\Big(\exp \Big(\int_0^1f(t)dW_t \Big)\mathbb{1}_{\{\int_0^1f(t)dW_t\ge2\}}\Big)$$

Do you have any idea how to tackle this? I don't expect full solution, I will appreciate any hint...

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Let

$$X_t := \int_0^t f(s)dW_s$$

Now $$Z_t:=\mathcal{E}(X)_t=\exp \left ( X_t - \frac{1}{2} [X]_t \right )=\exp \left ( \int_0^t f(s)dW_s - \frac{1}{2} \int_0^t f(s)^2ds \right )$$

Thus $$\mathbb{E}\Big(\exp \Big(\int_0^1f(t)dW_t \Big)\mathbb{1}_{\{\int_0^1f(t)dW_t\ge2\}}\Big)=\mathbb{E}\Big(Z_1\exp\Big(\frac{1}{2}\int_0^1f(s)^2ds\Big)\mathbb{1}_{\{\int_0^1f(t)dW_t\ge2\}}\Big)= $$ $$ \exp\Big(\frac{1}{2}\int_0^1f(s)^2ds\Big)\mathbb{P}_Q \Big(\Big\{\int_0^1f(t)dW_t\ge2 \Big\}\Big)$$

Our new Wiener process:

$$ \tilde W_t =W_t - \left [ W, X \right]_t=W_t-\int_0^tf(s)ds $$

So we have

$$\mathbb{P}_Q \Big(\Big\{\int_0^1f(t)dW_t\ge2 \Big\}\Big)=\mathbb{P}_Q \Big(\Big\{\int_0^1f(t)d\tilde W_t + \int_0^1 f(t)^2dt\ge2 \Big\}\Big)=\mathbb{P}_Q \Big(\Big\{\int_0^1f(t)d\tilde W_t \ge 0.1333 \Big\}\Big)$$

Answer 1

Apply Girsanov theorem with $X_t := \int_0^t f(t)dW_t$