How can I use the moment generating function to prove independence between $X_s$ and $X_t-X_s$












0












$begingroup$


Consider a process $X_t$ with the dynamics $$dX_t=f(t)dW_t$$
where $W_t$ is Brownian motion and $X_0=0$. Define $sigma^2(t)=int_0^tf^2(u)du$



Clearly $X_s$ and $X_t-X_s$ for $t>s$ are independent (due to the properties of Brownian motion) and $X_tsim (0,sigma^2(t))$



My question is: Can I use the moment generating function to prove the independence between $X_s$ and $X_t-X_s$?



Hence computing $E[exp( a_1X_s+a_2(X_t-X_s) )]$ and use the result to prove independence






probability stochastic-processes stochastic-calculus







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  • $begingroup$
    thanks! I have now corrected.
    $endgroup$
    – econmajorr
    Jan 16 at 18:35
















0












$begingroup$


Consider a process $X_t$ with the dynamics $$dX_t=f(t)dW_t$$
where $W_t$ is Brownian motion and $X_0=0$. Define $sigma^2(t)=int_0^tf^2(u)du$



Clearly $X_s$ and $X_t-X_s$ for $t>s$ are independent (due to the properties of Brownian motion) and $X_tsim (0,sigma^2(t))$



My question is: Can I use the moment generating function to prove the independence between $X_s$ and $X_t-X_s$?



Hence computing $E[exp( a_1X_s+a_2(X_t-X_s) )]$ and use the result to prove independence






probability stochastic-processes stochastic-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    thanks! I have now corrected.
    $endgroup$
    – econmajorr
    Jan 16 at 18:35














0












0








0





$begingroup$


Consider a process $X_t$ with the dynamics $$dX_t=f(t)dW_t$$
where $W_t$ is Brownian motion and $X_0=0$. Define $sigma^2(t)=int_0^tf^2(u)du$



Clearly $X_s$ and $X_t-X_s$ for $t>s$ are independent (due to the properties of Brownian motion) and $X_tsim (0,sigma^2(t))$



My question is: Can I use the moment generating function to prove the independence between $X_s$ and $X_t-X_s$?



Hence computing $E[exp( a_1X_s+a_2(X_t-X_s) )]$ and use the result to prove independence






probability stochastic-processes stochastic-calculus







share|cite|improve this question











$endgroup$




Consider a process $X_t$ with the dynamics $$dX_t=f(t)dW_t$$
where $W_t$ is Brownian motion and $X_0=0$. Define $sigma^2(t)=int_0^tf^2(u)du$



Clearly $X_s$ and $X_t-X_s$ for $t>s$ are independent (due to the properties of Brownian motion) and $X_tsim (0,sigma^2(t))$



My question is: Can I use the moment generating function to prove the independence between $X_s$ and $X_t-X_s$?



Hence computing $E[exp( a_1X_s+a_2(X_t-X_s) )]$ and use the result to prove independence






probability stochastic-processes stochastic-calculus




probability stochastic-processes stochastic-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 18:35







econmajorr

















asked Jan 16 at 17:44









econmajorreconmajorr

104




104












  • $begingroup$
    thanks! I have now corrected.
    $endgroup$
    – econmajorr
    Jan 16 at 18:35


















  • $begingroup$
    thanks! I have now corrected.
    $endgroup$
    – econmajorr
    Jan 16 at 18:35
















$begingroup$
thanks! I have now corrected.
$endgroup$
– econmajorr
Jan 16 at 18:35




$begingroup$
thanks! I have now corrected.
$endgroup$
– econmajorr
Jan 16 at 18:35










1 Answer
1






active

oldest

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0












$begingroup$

Let ${f^n}$ be a sequence of simple functions converging to $f$ and $X^n_ttriangleq int_0^tf^n(s)dW_s$. Then for each $n$, $X^n_s$ and $X^n_t-X^n_s$ are clearly independent; and we get the independence of the limits $X_s$ and $X_t-X_s$ from this result. Proof of the last result, as you can see, is likely to rely on characteristic functions as you desire.



Or you can repeat a step in Levy's characterization of BM (e.g., Karatzas and Shreve, Theorem 3.3.16). It suffices to show
$$
E(e^{iu(X_t-X_s)}|mathcal F_s)=e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}
$$
which is equivalent to the statement that for all $Ainmathcal F_s$,
$$
E(1_Ae^{iu(X_t-X_s)})=P(A)e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}.
$$
Now, apply Ito's lemma to $xmapsto e^{iux}$ to obtain
$$
e^{iuX_t}=e^{iuX_s}+iuint_s^te^{iuX_tau}f(tau)dW_tau-frac{1}{2}u^2int_s^te^{iuX_tau}f^2(tau)dtau.
$$
Under a mild regularity condition on $f$ (e.g., $int_0^tf^2(tau)dtau<infty$ for all $t$), the stochastic integral is a martingale, so
$$
E(e^{iuX_t}|mathcal F_s)=e^{iuX_s}-frac{1}{2}u^2int_s^tE(e^{iuX_tau}|mathcal F_s)f^2(tau)dtau.
$$
Multiply $1_Ae^{-iuX_s}$ and take expectations:
$$
E(1_Ae^{iu(X_t-X_s)})
=P(A)-frac{1}{2}u^2int_s^tE(1_Ae^{iu(X_tau-X_s)})f^2(tau)dtau.
$$
Let $z(t)=E(1_Ae^{iu(X_t-X_s)})$. We then have
$$
z(t)=P(A)-frac{1}{2}u^2int_s^tz(tau)f^2(tau)dtau,quad tge s,
$$
whose unique solution is
$$
z(t)=P(A)e^{-frac{1}{2}u^2int_s^tf(tau)^2dtau}.
$$






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    0












    $begingroup$

    Let ${f^n}$ be a sequence of simple functions converging to $f$ and $X^n_ttriangleq int_0^tf^n(s)dW_s$. Then for each $n$, $X^n_s$ and $X^n_t-X^n_s$ are clearly independent; and we get the independence of the limits $X_s$ and $X_t-X_s$ from this result. Proof of the last result, as you can see, is likely to rely on characteristic functions as you desire.



    Or you can repeat a step in Levy's characterization of BM (e.g., Karatzas and Shreve, Theorem 3.3.16). It suffices to show
    $$
    E(e^{iu(X_t-X_s)}|mathcal F_s)=e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}
    $$
    which is equivalent to the statement that for all $Ainmathcal F_s$,
    $$
    E(1_Ae^{iu(X_t-X_s)})=P(A)e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}.
    $$
    Now, apply Ito's lemma to $xmapsto e^{iux}$ to obtain
    $$
    e^{iuX_t}=e^{iuX_s}+iuint_s^te^{iuX_tau}f(tau)dW_tau-frac{1}{2}u^2int_s^te^{iuX_tau}f^2(tau)dtau.
    $$
    Under a mild regularity condition on $f$ (e.g., $int_0^tf^2(tau)dtau<infty$ for all $t$), the stochastic integral is a martingale, so
    $$
    E(e^{iuX_t}|mathcal F_s)=e^{iuX_s}-frac{1}{2}u^2int_s^tE(e^{iuX_tau}|mathcal F_s)f^2(tau)dtau.
    $$
    Multiply $1_Ae^{-iuX_s}$ and take expectations:
    $$
    E(1_Ae^{iu(X_t-X_s)})
    =P(A)-frac{1}{2}u^2int_s^tE(1_Ae^{iu(X_tau-X_s)})f^2(tau)dtau.
    $$
    Let $z(t)=E(1_Ae^{iu(X_t-X_s)})$. We then have
    $$
    z(t)=P(A)-frac{1}{2}u^2int_s^tz(tau)f^2(tau)dtau,quad tge s,
    $$
    whose unique solution is
    $$
    z(t)=P(A)e^{-frac{1}{2}u^2int_s^tf(tau)^2dtau}.
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let ${f^n}$ be a sequence of simple functions converging to $f$ and $X^n_ttriangleq int_0^tf^n(s)dW_s$. Then for each $n$, $X^n_s$ and $X^n_t-X^n_s$ are clearly independent; and we get the independence of the limits $X_s$ and $X_t-X_s$ from this result. Proof of the last result, as you can see, is likely to rely on characteristic functions as you desire.



      Or you can repeat a step in Levy's characterization of BM (e.g., Karatzas and Shreve, Theorem 3.3.16). It suffices to show
      $$
      E(e^{iu(X_t-X_s)}|mathcal F_s)=e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}
      $$
      which is equivalent to the statement that for all $Ainmathcal F_s$,
      $$
      E(1_Ae^{iu(X_t-X_s)})=P(A)e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}.
      $$
      Now, apply Ito's lemma to $xmapsto e^{iux}$ to obtain
      $$
      e^{iuX_t}=e^{iuX_s}+iuint_s^te^{iuX_tau}f(tau)dW_tau-frac{1}{2}u^2int_s^te^{iuX_tau}f^2(tau)dtau.
      $$
      Under a mild regularity condition on $f$ (e.g., $int_0^tf^2(tau)dtau<infty$ for all $t$), the stochastic integral is a martingale, so
      $$
      E(e^{iuX_t}|mathcal F_s)=e^{iuX_s}-frac{1}{2}u^2int_s^tE(e^{iuX_tau}|mathcal F_s)f^2(tau)dtau.
      $$
      Multiply $1_Ae^{-iuX_s}$ and take expectations:
      $$
      E(1_Ae^{iu(X_t-X_s)})
      =P(A)-frac{1}{2}u^2int_s^tE(1_Ae^{iu(X_tau-X_s)})f^2(tau)dtau.
      $$
      Let $z(t)=E(1_Ae^{iu(X_t-X_s)})$. We then have
      $$
      z(t)=P(A)-frac{1}{2}u^2int_s^tz(tau)f^2(tau)dtau,quad tge s,
      $$
      whose unique solution is
      $$
      z(t)=P(A)e^{-frac{1}{2}u^2int_s^tf(tau)^2dtau}.
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let ${f^n}$ be a sequence of simple functions converging to $f$ and $X^n_ttriangleq int_0^tf^n(s)dW_s$. Then for each $n$, $X^n_s$ and $X^n_t-X^n_s$ are clearly independent; and we get the independence of the limits $X_s$ and $X_t-X_s$ from this result. Proof of the last result, as you can see, is likely to rely on characteristic functions as you desire.



        Or you can repeat a step in Levy's characterization of BM (e.g., Karatzas and Shreve, Theorem 3.3.16). It suffices to show
        $$
        E(e^{iu(X_t-X_s)}|mathcal F_s)=e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}
        $$
        which is equivalent to the statement that for all $Ainmathcal F_s$,
        $$
        E(1_Ae^{iu(X_t-X_s)})=P(A)e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}.
        $$
        Now, apply Ito's lemma to $xmapsto e^{iux}$ to obtain
        $$
        e^{iuX_t}=e^{iuX_s}+iuint_s^te^{iuX_tau}f(tau)dW_tau-frac{1}{2}u^2int_s^te^{iuX_tau}f^2(tau)dtau.
        $$
        Under a mild regularity condition on $f$ (e.g., $int_0^tf^2(tau)dtau<infty$ for all $t$), the stochastic integral is a martingale, so
        $$
        E(e^{iuX_t}|mathcal F_s)=e^{iuX_s}-frac{1}{2}u^2int_s^tE(e^{iuX_tau}|mathcal F_s)f^2(tau)dtau.
        $$
        Multiply $1_Ae^{-iuX_s}$ and take expectations:
        $$
        E(1_Ae^{iu(X_t-X_s)})
        =P(A)-frac{1}{2}u^2int_s^tE(1_Ae^{iu(X_tau-X_s)})f^2(tau)dtau.
        $$
        Let $z(t)=E(1_Ae^{iu(X_t-X_s)})$. We then have
        $$
        z(t)=P(A)-frac{1}{2}u^2int_s^tz(tau)f^2(tau)dtau,quad tge s,
        $$
        whose unique solution is
        $$
        z(t)=P(A)e^{-frac{1}{2}u^2int_s^tf(tau)^2dtau}.
        $$






        share|cite|improve this answer









        $endgroup$



        Let ${f^n}$ be a sequence of simple functions converging to $f$ and $X^n_ttriangleq int_0^tf^n(s)dW_s$. Then for each $n$, $X^n_s$ and $X^n_t-X^n_s$ are clearly independent; and we get the independence of the limits $X_s$ and $X_t-X_s$ from this result. Proof of the last result, as you can see, is likely to rely on characteristic functions as you desire.



        Or you can repeat a step in Levy's characterization of BM (e.g., Karatzas and Shreve, Theorem 3.3.16). It suffices to show
        $$
        E(e^{iu(X_t-X_s)}|mathcal F_s)=e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}
        $$
        which is equivalent to the statement that for all $Ainmathcal F_s$,
        $$
        E(1_Ae^{iu(X_t-X_s)})=P(A)e^{-frac{1}{2}u^2int_s^tf^2(tau)dtau}.
        $$
        Now, apply Ito's lemma to $xmapsto e^{iux}$ to obtain
        $$
        e^{iuX_t}=e^{iuX_s}+iuint_s^te^{iuX_tau}f(tau)dW_tau-frac{1}{2}u^2int_s^te^{iuX_tau}f^2(tau)dtau.
        $$
        Under a mild regularity condition on $f$ (e.g., $int_0^tf^2(tau)dtau<infty$ for all $t$), the stochastic integral is a martingale, so
        $$
        E(e^{iuX_t}|mathcal F_s)=e^{iuX_s}-frac{1}{2}u^2int_s^tE(e^{iuX_tau}|mathcal F_s)f^2(tau)dtau.
        $$
        Multiply $1_Ae^{-iuX_s}$ and take expectations:
        $$
        E(1_Ae^{iu(X_t-X_s)})
        =P(A)-frac{1}{2}u^2int_s^tE(1_Ae^{iu(X_tau-X_s)})f^2(tau)dtau.
        $$
        Let $z(t)=E(1_Ae^{iu(X_t-X_s)})$. We then have
        $$
        z(t)=P(A)-frac{1}{2}u^2int_s^tz(tau)f^2(tau)dtau,quad tge s,
        $$
        whose unique solution is
        $$
        z(t)=P(A)e^{-frac{1}{2}u^2int_s^tf(tau)^2dtau}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 8:33









        AddSupAddSup

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