Compute $E(M_σ)$ when $σ = inf(t ≥ 0: |B_t| = 1)$ and $M_t = 4B^2_t +e^{4B_t−8t}−4t$












0












$begingroup$



Given $M_t = 4B^2_t +e^{4B_t−8t}−4t$ for $t ≥ 0$ and a Brownian motion $(B_t)_{t geq 0}$. Compute $E(M_σ)$ when $σ = inf(t ≥ 0: |B_t| = 1)$.




I have tried to show that $E|M_σ|leq K$ to apply Doob's Optional Sampling Theorem, thus $E(M_σ)=E(M_0)=1$, but my I am unsure how to explain the reasoning.



So far I've got $$E|M_σ|=E|M_{twedgeσ}|=E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}-8t|leq E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}|,$$ but don't know how to proceed. Any help would be greatly appreciated.






probability-theory brownian-motion martingales stopping-times







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  • $begingroup$
    "So far I have got $mathbb{E}|M_{sigma}| = mathbb{E}(|M_{t wedge sigma}|$) [...]" - how do you know that?
    $endgroup$
    – saz
    Jan 6 at 12:18












  • $begingroup$
    @saz I should have stated that we knew $M_t$ is a martingale. But thank you for your help
    $endgroup$
    – Zugzwangerz
    Jan 6 at 12:41










  • $begingroup$
    I'm well aware that $M$ is a martingale but $(|M_t|)_{t geq 0}$ is not a martingale. Moreover, we cannot directly apply the optional stopping theorem to $sigma$.
    $endgroup$
    – saz
    Jan 6 at 12:44
















0












$begingroup$



Given $M_t = 4B^2_t +e^{4B_t−8t}−4t$ for $t ≥ 0$ and a Brownian motion $(B_t)_{t geq 0}$. Compute $E(M_σ)$ when $σ = inf(t ≥ 0: |B_t| = 1)$.




I have tried to show that $E|M_σ|leq K$ to apply Doob's Optional Sampling Theorem, thus $E(M_σ)=E(M_0)=1$, but my I am unsure how to explain the reasoning.



So far I've got $$E|M_σ|=E|M_{twedgeσ}|=E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}-8t|leq E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}|,$$ but don't know how to proceed. Any help would be greatly appreciated.






probability-theory brownian-motion martingales stopping-times







share|cite|improve this question











$endgroup$












  • $begingroup$
    "So far I have got $mathbb{E}|M_{sigma}| = mathbb{E}(|M_{t wedge sigma}|$) [...]" - how do you know that?
    $endgroup$
    – saz
    Jan 6 at 12:18












  • $begingroup$
    @saz I should have stated that we knew $M_t$ is a martingale. But thank you for your help
    $endgroup$
    – Zugzwangerz
    Jan 6 at 12:41










  • $begingroup$
    I'm well aware that $M$ is a martingale but $(|M_t|)_{t geq 0}$ is not a martingale. Moreover, we cannot directly apply the optional stopping theorem to $sigma$.
    $endgroup$
    – saz
    Jan 6 at 12:44














0












0








0





$begingroup$



Given $M_t = 4B^2_t +e^{4B_t−8t}−4t$ for $t ≥ 0$ and a Brownian motion $(B_t)_{t geq 0}$. Compute $E(M_σ)$ when $σ = inf(t ≥ 0: |B_t| = 1)$.




I have tried to show that $E|M_σ|leq K$ to apply Doob's Optional Sampling Theorem, thus $E(M_σ)=E(M_0)=1$, but my I am unsure how to explain the reasoning.



So far I've got $$E|M_σ|=E|M_{twedgeσ}|=E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}-8t|leq E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}|,$$ but don't know how to proceed. Any help would be greatly appreciated.






probability-theory brownian-motion martingales stopping-times







share|cite|improve this question











$endgroup$





Given $M_t = 4B^2_t +e^{4B_t−8t}−4t$ for $t ≥ 0$ and a Brownian motion $(B_t)_{t geq 0}$. Compute $E(M_σ)$ when $σ = inf(t ≥ 0: |B_t| = 1)$.




I have tried to show that $E|M_σ|leq K$ to apply Doob's Optional Sampling Theorem, thus $E(M_σ)=E(M_0)=1$, but my I am unsure how to explain the reasoning.



So far I've got $$E|M_σ|=E|M_{twedgeσ}|=E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}-8t|leq E|4B_{twedgeσ}^2+e^{4B_{twedgeσ}-8{twedgeσ}}|,$$ but don't know how to proceed. Any help would be greatly appreciated.






probability-theory brownian-motion martingales stopping-times




probability-theory brownian-motion martingales stopping-times






share|cite|improve this question















share|cite|improve this question













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edited Jan 6 at 12:44









Did

248k23226465




248k23226465










asked Jan 6 at 11:44









ZugzwangerzZugzwangerz

223




223












  • $begingroup$
    "So far I have got $mathbb{E}|M_{sigma}| = mathbb{E}(|M_{t wedge sigma}|$) [...]" - how do you know that?
    $endgroup$
    – saz
    Jan 6 at 12:18












  • $begingroup$
    @saz I should have stated that we knew $M_t$ is a martingale. But thank you for your help
    $endgroup$
    – Zugzwangerz
    Jan 6 at 12:41










  • $begingroup$
    I'm well aware that $M$ is a martingale but $(|M_t|)_{t geq 0}$ is not a martingale. Moreover, we cannot directly apply the optional stopping theorem to $sigma$.
    $endgroup$
    – saz
    Jan 6 at 12:44


















  • $begingroup$
    "So far I have got $mathbb{E}|M_{sigma}| = mathbb{E}(|M_{t wedge sigma}|$) [...]" - how do you know that?
    $endgroup$
    – saz
    Jan 6 at 12:18












  • $begingroup$
    @saz I should have stated that we knew $M_t$ is a martingale. But thank you for your help
    $endgroup$
    – Zugzwangerz
    Jan 6 at 12:41










  • $begingroup$
    I'm well aware that $M$ is a martingale but $(|M_t|)_{t geq 0}$ is not a martingale. Moreover, we cannot directly apply the optional stopping theorem to $sigma$.
    $endgroup$
    – saz
    Jan 6 at 12:44
















$begingroup$
"So far I have got $mathbb{E}|M_{sigma}| = mathbb{E}(|M_{t wedge sigma}|$) [...]" - how do you know that?
$endgroup$
– saz
Jan 6 at 12:18






$begingroup$
"So far I have got $mathbb{E}|M_{sigma}| = mathbb{E}(|M_{t wedge sigma}|$) [...]" - how do you know that?
$endgroup$
– saz
Jan 6 at 12:18














$begingroup$
@saz I should have stated that we knew $M_t$ is a martingale. But thank you for your help
$endgroup$
– Zugzwangerz
Jan 6 at 12:41




$begingroup$
@saz I should have stated that we knew $M_t$ is a martingale. But thank you for your help
$endgroup$
– Zugzwangerz
Jan 6 at 12:41












$begingroup$
I'm well aware that $M$ is a martingale but $(|M_t|)_{t geq 0}$ is not a martingale. Moreover, we cannot directly apply the optional stopping theorem to $sigma$.
$endgroup$
– saz
Jan 6 at 12:44




$begingroup$
I'm well aware that $M$ is a martingale but $(|M_t|)_{t geq 0}$ is not a martingale. Moreover, we cannot directly apply the optional stopping theorem to $sigma$.
$endgroup$
– saz
Jan 6 at 12:44










1 Answer
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$begingroup$

Note that $M_t = X_t+Y_t$ with $$X_t := 4(B_t^2-t) qquad Y_t := e^{4B_t-8t}.$$ In particular, $mathbb{E}(M_{sigma}) = mathbb{E}(X_{sigma})+mathbb{E}(Y_{sigma})$, and therefore it suffices to calculate $mathbb{E}(X_{sigma})$ and $mathbb{E}(Y_{sigma})$.



Calculation of $mathbb{E}(X_{sigma})$:




  1. Check (or recall) that $(X_t)_{t geq 0}$ is a martingale.

  2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge sigma})=0,$$ i.e. $$mathbb{E}(B_{t wedge sigma}^2) = mathbb{E}(t wedge sigma).$$

  3. Show that $|B_{t wedge sigma}| leq 1$ for all $t geq 0$.

  4. Use the dominated convergence theorem and the monotone convergence theorem to deduce from Step 2&3 that $$mathbb{E}(B_{sigma}^2) = mathbb{E}(sigma),$$ i.e. $$mathbb{E}(X_{sigma})=0.$$


Calculation of $mathbb{E}(Y_{sigma})$:




  1. Check that $(Y_t)_{t geq 0}$ is a martingale.

  2. Apply the optional stopping theorem to show that $mathbb{E}(Y_{t wedge sigma})=1$ for all $t geq 0$.

  3. Conclude from $|B_{t wedge sigma}| leq 1$ that $|Y_{t wedge sigma}| leq e^{4}$ for all $t geq 0$.

  4. Apply the dominated convergence theorem to deduce from Step 2& 3 that $mathbb{E}(Y_{sigma})=1$.






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    $begingroup$

    Note that $M_t = X_t+Y_t$ with $$X_t := 4(B_t^2-t) qquad Y_t := e^{4B_t-8t}.$$ In particular, $mathbb{E}(M_{sigma}) = mathbb{E}(X_{sigma})+mathbb{E}(Y_{sigma})$, and therefore it suffices to calculate $mathbb{E}(X_{sigma})$ and $mathbb{E}(Y_{sigma})$.



    Calculation of $mathbb{E}(X_{sigma})$:




    1. Check (or recall) that $(X_t)_{t geq 0}$ is a martingale.

    2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge sigma})=0,$$ i.e. $$mathbb{E}(B_{t wedge sigma}^2) = mathbb{E}(t wedge sigma).$$

    3. Show that $|B_{t wedge sigma}| leq 1$ for all $t geq 0$.

    4. Use the dominated convergence theorem and the monotone convergence theorem to deduce from Step 2&3 that $$mathbb{E}(B_{sigma}^2) = mathbb{E}(sigma),$$ i.e. $$mathbb{E}(X_{sigma})=0.$$


    Calculation of $mathbb{E}(Y_{sigma})$:




    1. Check that $(Y_t)_{t geq 0}$ is a martingale.

    2. Apply the optional stopping theorem to show that $mathbb{E}(Y_{t wedge sigma})=1$ for all $t geq 0$.

    3. Conclude from $|B_{t wedge sigma}| leq 1$ that $|Y_{t wedge sigma}| leq e^{4}$ for all $t geq 0$.

    4. Apply the dominated convergence theorem to deduce from Step 2& 3 that $mathbb{E}(Y_{sigma})=1$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Note that $M_t = X_t+Y_t$ with $$X_t := 4(B_t^2-t) qquad Y_t := e^{4B_t-8t}.$$ In particular, $mathbb{E}(M_{sigma}) = mathbb{E}(X_{sigma})+mathbb{E}(Y_{sigma})$, and therefore it suffices to calculate $mathbb{E}(X_{sigma})$ and $mathbb{E}(Y_{sigma})$.



      Calculation of $mathbb{E}(X_{sigma})$:




      1. Check (or recall) that $(X_t)_{t geq 0}$ is a martingale.

      2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge sigma})=0,$$ i.e. $$mathbb{E}(B_{t wedge sigma}^2) = mathbb{E}(t wedge sigma).$$

      3. Show that $|B_{t wedge sigma}| leq 1$ for all $t geq 0$.

      4. Use the dominated convergence theorem and the monotone convergence theorem to deduce from Step 2&3 that $$mathbb{E}(B_{sigma}^2) = mathbb{E}(sigma),$$ i.e. $$mathbb{E}(X_{sigma})=0.$$


      Calculation of $mathbb{E}(Y_{sigma})$:




      1. Check that $(Y_t)_{t geq 0}$ is a martingale.

      2. Apply the optional stopping theorem to show that $mathbb{E}(Y_{t wedge sigma})=1$ for all $t geq 0$.

      3. Conclude from $|B_{t wedge sigma}| leq 1$ that $|Y_{t wedge sigma}| leq e^{4}$ for all $t geq 0$.

      4. Apply the dominated convergence theorem to deduce from Step 2& 3 that $mathbb{E}(Y_{sigma})=1$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Note that $M_t = X_t+Y_t$ with $$X_t := 4(B_t^2-t) qquad Y_t := e^{4B_t-8t}.$$ In particular, $mathbb{E}(M_{sigma}) = mathbb{E}(X_{sigma})+mathbb{E}(Y_{sigma})$, and therefore it suffices to calculate $mathbb{E}(X_{sigma})$ and $mathbb{E}(Y_{sigma})$.



        Calculation of $mathbb{E}(X_{sigma})$:




        1. Check (or recall) that $(X_t)_{t geq 0}$ is a martingale.

        2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge sigma})=0,$$ i.e. $$mathbb{E}(B_{t wedge sigma}^2) = mathbb{E}(t wedge sigma).$$

        3. Show that $|B_{t wedge sigma}| leq 1$ for all $t geq 0$.

        4. Use the dominated convergence theorem and the monotone convergence theorem to deduce from Step 2&3 that $$mathbb{E}(B_{sigma}^2) = mathbb{E}(sigma),$$ i.e. $$mathbb{E}(X_{sigma})=0.$$


        Calculation of $mathbb{E}(Y_{sigma})$:




        1. Check that $(Y_t)_{t geq 0}$ is a martingale.

        2. Apply the optional stopping theorem to show that $mathbb{E}(Y_{t wedge sigma})=1$ for all $t geq 0$.

        3. Conclude from $|B_{t wedge sigma}| leq 1$ that $|Y_{t wedge sigma}| leq e^{4}$ for all $t geq 0$.

        4. Apply the dominated convergence theorem to deduce from Step 2& 3 that $mathbb{E}(Y_{sigma})=1$.






        share|cite|improve this answer











        $endgroup$



        Note that $M_t = X_t+Y_t$ with $$X_t := 4(B_t^2-t) qquad Y_t := e^{4B_t-8t}.$$ In particular, $mathbb{E}(M_{sigma}) = mathbb{E}(X_{sigma})+mathbb{E}(Y_{sigma})$, and therefore it suffices to calculate $mathbb{E}(X_{sigma})$ and $mathbb{E}(Y_{sigma})$.



        Calculation of $mathbb{E}(X_{sigma})$:




        1. Check (or recall) that $(X_t)_{t geq 0}$ is a martingale.

        2. Apply the optional stopping theorem to show that $$mathbb{E}(X_{t wedge sigma})=0,$$ i.e. $$mathbb{E}(B_{t wedge sigma}^2) = mathbb{E}(t wedge sigma).$$

        3. Show that $|B_{t wedge sigma}| leq 1$ for all $t geq 0$.

        4. Use the dominated convergence theorem and the monotone convergence theorem to deduce from Step 2&3 that $$mathbb{E}(B_{sigma}^2) = mathbb{E}(sigma),$$ i.e. $$mathbb{E}(X_{sigma})=0.$$


        Calculation of $mathbb{E}(Y_{sigma})$:




        1. Check that $(Y_t)_{t geq 0}$ is a martingale.

        2. Apply the optional stopping theorem to show that $mathbb{E}(Y_{t wedge sigma})=1$ for all $t geq 0$.

        3. Conclude from $|B_{t wedge sigma}| leq 1$ that $|Y_{t wedge sigma}| leq e^{4}$ for all $t geq 0$.

        4. Apply the dominated convergence theorem to deduce from Step 2& 3 that $mathbb{E}(Y_{sigma})=1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 14:01

























        answered Jan 6 at 12:27









        sazsaz

        81.8k862130




        81.8k862130






























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