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We have $X(t)=[X_1(t)\ X_2(t)\ X_3(t)\ \dots\ X_n(t)]$ and $Y(t)=[Y_1(t)\ Y_2(t)\ Y_3(t)\ \dots\ Y_n(t)]$ are two stochastic process such that: $$\sup E[Y_1^2] \leq K,$$ on $[t_0, T]$ with $K$ a positive integer.

Is that true

$$E\int^T_{t_0} Y_1^{2} F(X(t))^2 \, dt \leq \sup E[Y_1^2]E\int^T_{t_0} F(X(t))^2 \, dt$$

Important note: $Y_1(t)$ is a function of $X_1$.

Michael Hardy
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fidel
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1 Answers1

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In general I don't see any reason why this should be true. However, if $Y_1(t)^2$ and $F(X(t))$ are independent for Lebesgue a.e. $t$ (it would be sufficient to assume $Y_1$ and $X$ are independent, of course), then it does hold.

You may say by Tonelli's Theorem (Fubini's Theorem for nonnegative integrands) $$ E\int_{t_0}^T Y_1(t)^2 F(X(t))^2~dt = \int_{t_0}^T E[Y_1(t)^2 F(X(t))^2]~dt $$ which by independence is $$ =\int_{t_0}^T E[Y_1(t)^2] E[F(X(t))^2]~dt $$ and by monotonicity of the integral is $$ \leq \int_{t_0}^T (\sup_s E[Y_1(s)^2]) E[F(X(t))^2]~dt $$ $$ = \sup_t E[Y_1(t)^2]\int_{t_0}^T E[F(X(t))^2]~dt $$ and again by Tonelli's Theorem this is $$ =\sup_t E[Y_1(t)^2]E\int_{t_0}^T F(X(t))^2~dt. $$

nullUser
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  • the reality i was trying to prove that $$E\int^T_{t_0} Y_1^{2} F(X(t))^2 , dt < infinity , $$ but because Y1(t) and X1(t) are dependent so i didn't found a manner to prove it , i have also supE[F(X(t))^2]≤C, – fidel Jun 05 '13 at 12:24
  • because if i ensure $$E\int^T_{t_0} Y_1^{2} F(X(t))^2 , dt \leq infinity , $$ By the ito property i can deduce that $$E\int^T_{t_0} Y_1 F(X(t))dw , =0 , $$ after a deep reading i think that i can use Holder's inequality to prove:$$E\int^T_{t_0} Y_1^{2} F(X(t))^2 , dt \leq infinity , $$ even X1(t) and Y1(t) are dependent – fidel Jun 05 '13 at 14:14
  • In the future please only ask questions that you want to know the answer to. – nullUser Jun 05 '13 at 14:26
  • i just asked a part of a question because i think that if this part will be resolved the remainder is easy to do, but unfortunately this part is so hard to do. any way i'm so sorry if i made you confused. – fidel Jun 05 '13 at 14:35
  • but i realy appriciate your help teacher. – fidel Jun 05 '13 at 14:37
  • this is my question,(which i want to know the answer), we have $$\sup E[Y_1^2] \leq K,$$ on $[t_0, T]$ with $K$ a positive integer. and $$\sup E[F(X(t))_^2] \leq C,$$ on $[t_0, T]$ with $C$ a positive integer. i want to know if $$E\int^T_{t_0} Y_1^{2} F(X(t))^2 , dt \leq <infinity, $$ thus, $$E\int^T_{t_0} Y_1 F(X(t)) , dW , $$ is an ito process – fidel Jun 05 '13 at 14:40
  • this is my question,(which i want to know the answer), we have $$\sup E[Y_1^2] \leq K,$$ on $[t_0, T]$ with $K$ a positive integer. and $$\sup E[F(X(t))^2] \leq C,$$ on $[t_0, T]$ with $C$ a positive integer. i want to know if $$E\int^T_{t_0} Y_1^{2} F(X(t))^2 , dt < infinity, $$ thus, $$E\int^T_{t_0} Y_1 F(X(t)) , dW , $$ is an ito process – fidel Jun 05 '13 at 14:47
  • You need to make some assumptions on $F$ or on $F(X)$. Since you want to pull out the $Y_1^2$, the only way you could do this with Holder's inequality, without assuming independence, and get something finite would be if you assumed $F(X)$ is essentially bounded for a.e. $t$. – nullUser Jun 05 '13 at 15:26
  • i think even with holder's inequality it still difficult, because holder's inequality leads only to show that $$E\int^T_{t_0} Y_1 F(X(t)) , dt <infinity, $$ and not $$E\int^T_{t_0} Y_1^{2} F(X(t))^2 , dt <infinity, $$ – fidel Jun 05 '13 at 15:48
  • i will look about another solution, thanks teacher, so sorry for troubling you – fidel Jun 05 '13 at 15:58