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Fix a random variable $X$ (measurable at time zero if you like) which is finite but not in $L^\infty$, then define $\lambda_t = X$ for all $t$. Then $$ \int_0^T \lambda_t^2\,dt = T X^2. $$ By assumption $T X^2 < \infty$ for all $T, \omega$, but there is no $C$ such that $TX^2 \leq C$ almost surely, as this would imply $X \in L^\infty$. The difference ...


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Using summation by parts, as suggested, it actually becomes quite clear. $$\sum_{n=1}^{N}t_n(W(t_{n+1})-W(t_n))=t_N W(t_N)-t_1W(t_1)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ With $t_N=T$ and $t_1=0$. This gives: $$\int_0^TtdW(t)= TW(T)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ Taking $n \rightarrow \infty$ gives the Riemann integral. Thus the results ends up being: ...


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Hint: Use the summation by parts formula (also known as Abel's summation formula): $$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_{n+1} g_{n+1}-f_m g_m) - \sum_{k=m}^n g_{k+1} (f_{k+1}-f_k).$$


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It should be Lebesgue integral, instead of Riemann integral. See https://almostsure.wordpress.com/2010/06/02/failure-of-pathwise-integration-for-fv-processes/#more-659 Second paragraph after Lemma 1, it is specifically stated: it can be shown that if V is a continuous FV process, then a predictable process {\xi} is V-integrable if and only if it is ...


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I belive we should mention using Fubini's theorem to swap integral and expectation signs: $\mathbb{E}\left[\int^T_0 W_t^2 dt\right]=\int^T_0\mathbb{E}[W^2_t]dt=\frac{T^2}{2}$


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$$\frac{dx_t}{x_t}=\mu+\sigma dz_t\\$$let $f(x)=ln(x)$ apply ito formula to find $df(x)$ $$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac{1}{2}\frac{\partial^2f }{\partial x^2}\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\$$now put $dx_t=\mu x_tdt+\sigma x_t dz_t$ ...


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Thank you for the clarification. From step 1 I've got: $$Y(t)=\exp(-\frac{5}{2}t + B(t)) \ \ \ \ , \ \ \ \ Y(0)=1$$ so per my understanding we can assume that $$X(t)=Y(t)V(t)$$ where $$dY(t)=\beta(t)Y(t) dt + \delta(t)Y(t) dB(t)$$ and $$dV(t)=a(t) dt + b(t) dB(t) \ \ \ \ \ \ \ , \ \ \ \ \ V(0)=X(0)=0$$ now: $$dX(t)=Y(t)dV(t) + V(t) dY(t) + dV(t) dY(t)$$ ...


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Hints Solve the homogeneous linear SDE $$dY_t = -2 Y_t \, dt + Y_t \, dB_t, \quad Y_0 = y_0 \tag{1}$$ To do so, use Itô's formula for $Z_t := \ln Y_t$. Let $X_t^0$ such that $\frac{1}{X_t^0}$ solves $(1)$ with initial condition $X_0^0=1$. Calculate $dX_t^0$ using step 1. Now apply Itô's formula to $Z_t := X_t \cdot X_t^0$. This gives you an integral ...



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