Conditional expectation of exponent to the power of BM, representation theorem












1












$begingroup$


Let $T > 0$ and $M(t) = E[e^{W(T)}|F(t)]$ where ${F(t) : t geq 0}$ is a natural filtration generated by W and $t leq T$
I need to show that M(t) is a martingale and I also need to find a unique square integrable function $phi$ for the representation
$M(t) = M(0) + int_0^t phi(s)dW(s)$




What I did:
for the martingale part I checked 3 conditions:




  1. It does look adapted, doesn't depend on future values (not sure how to prove this rigorously)

  2. Integrability: $E[|E[e^{W(T)}|F(t)]|] leq (jensen) E[E[e^{|W(T)|}|F(t)]] = E[e^{|W(T)|}] = E[e^{sqrt{T}*|epsilon|}] < infty $ (I can just say that MGF is finite, right? $epsilon$ is a standard normal random variable)


  3. For any $s leq t: E[M(t)|F(s)] = E[E[e^{W(T)}|F(t)]|F(s)] = E[E[e^{W(T)}|F(s)]|F(t)] = E[e^{W(T)}|F(s)] = M(s) $
    Thus $M(t)$ is a martingale



For the representation part:



$E[e^{W(T)}|F(t)] = M(0) + int_0^t phi(s)dW(s)$



Now, this is where I'm not sure if I'm doing it right:



$E[e^{W(T)}|F(t)] = E[e^{W(T) - W(t) + W(t)}|F(t)] = $ (take out what is known) $ e^{W(t)}*E[e^{W(T) - W(t)}|F(t)] = $ (independent increments) $ e^{W(t)}*E[e^{W(T)-W(t)}] = e^{W(t)}*e^{frac{T-t}{2}} = e^{W(t) + 0.5(T - t)} $



$M(0) = e^{0.5T}$, so my representation turns into this:



$e^{W(t) - 0.5t} = int_0^t phi(s)dW(s)$



Now, I am going to use Ito on a function $f(x,t) = e^{x - 0.5t}$



$f_x = e^{x - 0.5t}$



$f_{xx} = e^{x - 0.5t}$



$f_t = -0.5e^{x - 0.5t}$



$d(f) = e^{x - 0.5t}dx - 0.5e^{x - 0.5t}dt + 0.5*e^{x - 0.5t}dt = e^{x - 0.5t}dx$



$e^{W(t) - 0.5t} = 1 + int_0^t e^{W(s)-0.5s}dW(s) = int_0^t frac{1}{W(t)} + e^{W(s) - 0.5s}dW(s)$



So my $phi$ is $frac{1}{W(t)} + e^{W(s) - 0.5s}$



I need to check that it is square-integrable, but for some reason I'm confused about the $frac{1}{W(t)}$. Do I treat it as a constant and freely take out of expectation or ... maybe I made a mistake somewhere?
I hope someone can give me a helping hand.






stochastic-processes stochastic-calculus martingales stochastic-integrals stochastic-analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $T > 0$ and $M(t) = E[e^{W(T)}|F(t)]$ where ${F(t) : t geq 0}$ is a natural filtration generated by W and $t leq T$
    I need to show that M(t) is a martingale and I also need to find a unique square integrable function $phi$ for the representation
    $M(t) = M(0) + int_0^t phi(s)dW(s)$




    What I did:
    for the martingale part I checked 3 conditions:




    1. It does look adapted, doesn't depend on future values (not sure how to prove this rigorously)

    2. Integrability: $E[|E[e^{W(T)}|F(t)]|] leq (jensen) E[E[e^{|W(T)|}|F(t)]] = E[e^{|W(T)|}] = E[e^{sqrt{T}*|epsilon|}] < infty $ (I can just say that MGF is finite, right? $epsilon$ is a standard normal random variable)


    3. For any $s leq t: E[M(t)|F(s)] = E[E[e^{W(T)}|F(t)]|F(s)] = E[E[e^{W(T)}|F(s)]|F(t)] = E[e^{W(T)}|F(s)] = M(s) $
      Thus $M(t)$ is a martingale



    For the representation part:



    $E[e^{W(T)}|F(t)] = M(0) + int_0^t phi(s)dW(s)$



    Now, this is where I'm not sure if I'm doing it right:



    $E[e^{W(T)}|F(t)] = E[e^{W(T) - W(t) + W(t)}|F(t)] = $ (take out what is known) $ e^{W(t)}*E[e^{W(T) - W(t)}|F(t)] = $ (independent increments) $ e^{W(t)}*E[e^{W(T)-W(t)}] = e^{W(t)}*e^{frac{T-t}{2}} = e^{W(t) + 0.5(T - t)} $



    $M(0) = e^{0.5T}$, so my representation turns into this:



    $e^{W(t) - 0.5t} = int_0^t phi(s)dW(s)$



    Now, I am going to use Ito on a function $f(x,t) = e^{x - 0.5t}$



    $f_x = e^{x - 0.5t}$



    $f_{xx} = e^{x - 0.5t}$



    $f_t = -0.5e^{x - 0.5t}$



    $d(f) = e^{x - 0.5t}dx - 0.5e^{x - 0.5t}dt + 0.5*e^{x - 0.5t}dt = e^{x - 0.5t}dx$



    $e^{W(t) - 0.5t} = 1 + int_0^t e^{W(s)-0.5s}dW(s) = int_0^t frac{1}{W(t)} + e^{W(s) - 0.5s}dW(s)$



    So my $phi$ is $frac{1}{W(t)} + e^{W(s) - 0.5s}$



    I need to check that it is square-integrable, but for some reason I'm confused about the $frac{1}{W(t)}$. Do I treat it as a constant and freely take out of expectation or ... maybe I made a mistake somewhere?
    I hope someone can give me a helping hand.






    stochastic-processes stochastic-calculus martingales stochastic-integrals stochastic-analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $T > 0$ and $M(t) = E[e^{W(T)}|F(t)]$ where ${F(t) : t geq 0}$ is a natural filtration generated by W and $t leq T$
      I need to show that M(t) is a martingale and I also need to find a unique square integrable function $phi$ for the representation
      $M(t) = M(0) + int_0^t phi(s)dW(s)$




      What I did:
      for the martingale part I checked 3 conditions:




      1. It does look adapted, doesn't depend on future values (not sure how to prove this rigorously)

      2. Integrability: $E[|E[e^{W(T)}|F(t)]|] leq (jensen) E[E[e^{|W(T)|}|F(t)]] = E[e^{|W(T)|}] = E[e^{sqrt{T}*|epsilon|}] < infty $ (I can just say that MGF is finite, right? $epsilon$ is a standard normal random variable)


      3. For any $s leq t: E[M(t)|F(s)] = E[E[e^{W(T)}|F(t)]|F(s)] = E[E[e^{W(T)}|F(s)]|F(t)] = E[e^{W(T)}|F(s)] = M(s) $
        Thus $M(t)$ is a martingale



      For the representation part:



      $E[e^{W(T)}|F(t)] = M(0) + int_0^t phi(s)dW(s)$



      Now, this is where I'm not sure if I'm doing it right:



      $E[e^{W(T)}|F(t)] = E[e^{W(T) - W(t) + W(t)}|F(t)] = $ (take out what is known) $ e^{W(t)}*E[e^{W(T) - W(t)}|F(t)] = $ (independent increments) $ e^{W(t)}*E[e^{W(T)-W(t)}] = e^{W(t)}*e^{frac{T-t}{2}} = e^{W(t) + 0.5(T - t)} $



      $M(0) = e^{0.5T}$, so my representation turns into this:



      $e^{W(t) - 0.5t} = int_0^t phi(s)dW(s)$



      Now, I am going to use Ito on a function $f(x,t) = e^{x - 0.5t}$



      $f_x = e^{x - 0.5t}$



      $f_{xx} = e^{x - 0.5t}$



      $f_t = -0.5e^{x - 0.5t}$



      $d(f) = e^{x - 0.5t}dx - 0.5e^{x - 0.5t}dt + 0.5*e^{x - 0.5t}dt = e^{x - 0.5t}dx$



      $e^{W(t) - 0.5t} = 1 + int_0^t e^{W(s)-0.5s}dW(s) = int_0^t frac{1}{W(t)} + e^{W(s) - 0.5s}dW(s)$



      So my $phi$ is $frac{1}{W(t)} + e^{W(s) - 0.5s}$



      I need to check that it is square-integrable, but for some reason I'm confused about the $frac{1}{W(t)}$. Do I treat it as a constant and freely take out of expectation or ... maybe I made a mistake somewhere?
      I hope someone can give me a helping hand.






      stochastic-processes stochastic-calculus martingales stochastic-integrals stochastic-analysis







      share|cite|improve this question











      $endgroup$




      Let $T > 0$ and $M(t) = E[e^{W(T)}|F(t)]$ where ${F(t) : t geq 0}$ is a natural filtration generated by W and $t leq T$
      I need to show that M(t) is a martingale and I also need to find a unique square integrable function $phi$ for the representation
      $M(t) = M(0) + int_0^t phi(s)dW(s)$




      What I did:
      for the martingale part I checked 3 conditions:




      1. It does look adapted, doesn't depend on future values (not sure how to prove this rigorously)

      2. Integrability: $E[|E[e^{W(T)}|F(t)]|] leq (jensen) E[E[e^{|W(T)|}|F(t)]] = E[e^{|W(T)|}] = E[e^{sqrt{T}*|epsilon|}] < infty $ (I can just say that MGF is finite, right? $epsilon$ is a standard normal random variable)


      3. For any $s leq t: E[M(t)|F(s)] = E[E[e^{W(T)}|F(t)]|F(s)] = E[E[e^{W(T)}|F(s)]|F(t)] = E[e^{W(T)}|F(s)] = M(s) $
        Thus $M(t)$ is a martingale



      For the representation part:



      $E[e^{W(T)}|F(t)] = M(0) + int_0^t phi(s)dW(s)$



      Now, this is where I'm not sure if I'm doing it right:



      $E[e^{W(T)}|F(t)] = E[e^{W(T) - W(t) + W(t)}|F(t)] = $ (take out what is known) $ e^{W(t)}*E[e^{W(T) - W(t)}|F(t)] = $ (independent increments) $ e^{W(t)}*E[e^{W(T)-W(t)}] = e^{W(t)}*e^{frac{T-t}{2}} = e^{W(t) + 0.5(T - t)} $



      $M(0) = e^{0.5T}$, so my representation turns into this:



      $e^{W(t) - 0.5t} = int_0^t phi(s)dW(s)$



      Now, I am going to use Ito on a function $f(x,t) = e^{x - 0.5t}$



      $f_x = e^{x - 0.5t}$



      $f_{xx} = e^{x - 0.5t}$



      $f_t = -0.5e^{x - 0.5t}$



      $d(f) = e^{x - 0.5t}dx - 0.5e^{x - 0.5t}dt + 0.5*e^{x - 0.5t}dt = e^{x - 0.5t}dx$



      $e^{W(t) - 0.5t} = 1 + int_0^t e^{W(s)-0.5s}dW(s) = int_0^t frac{1}{W(t)} + e^{W(s) - 0.5s}dW(s)$



      So my $phi$ is $frac{1}{W(t)} + e^{W(s) - 0.5s}$



      I need to check that it is square-integrable, but for some reason I'm confused about the $frac{1}{W(t)}$. Do I treat it as a constant and freely take out of expectation or ... maybe I made a mistake somewhere?
      I hope someone can give me a helping hand.






      stochastic-processes stochastic-calculus martingales stochastic-integrals stochastic-analysis




      stochastic-processes stochastic-calculus martingales stochastic-integrals stochastic-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 6:23







      Makina

















      asked Nov 28 '18 at 5:58









      MakinaMakina

      1,1581316




      1,1581316






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          You already applied Ito's lemma to find
          $$
          e^{W(T)-T/2}=1+int_{0}^{T}e^{W(s)-s/2}dW(s).
          $$
          Next, note that
          begin{multline*}
          mathbb{E}left[int_{0}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]=int_{0}^{t}e^{W(s)-s/2}dW(s)+mathbb{E}left[int_{t}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]\=int_{0}^{t}e^{W(s)-s/2}dW(s).
          end{multline*}
          Therefore, applying conditional expectations to the first equation,
          $$
          e^{-T/2}M(t)=1+int_{0}^{t}e^{W(s)-s/2}dW(s).
          $$
          Multiplying by $e^{T/2}$ yields the desired result.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
            $endgroup$
            – Makina
            Nov 28 '18 at 6:41












          • $begingroup$
            I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
            $endgroup$
            – parsiad
            Dec 2 '18 at 3:32










          • $begingroup$
            Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
            $endgroup$
            – Makina
            Dec 2 '18 at 3:38











          Your Answer





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          1 Answer
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          1 Answer
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          active

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          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          You already applied Ito's lemma to find
          $$
          e^{W(T)-T/2}=1+int_{0}^{T}e^{W(s)-s/2}dW(s).
          $$
          Next, note that
          begin{multline*}
          mathbb{E}left[int_{0}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]=int_{0}^{t}e^{W(s)-s/2}dW(s)+mathbb{E}left[int_{t}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]\=int_{0}^{t}e^{W(s)-s/2}dW(s).
          end{multline*}
          Therefore, applying conditional expectations to the first equation,
          $$
          e^{-T/2}M(t)=1+int_{0}^{t}e^{W(s)-s/2}dW(s).
          $$
          Multiplying by $e^{T/2}$ yields the desired result.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
            $endgroup$
            – Makina
            Nov 28 '18 at 6:41












          • $begingroup$
            I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
            $endgroup$
            – parsiad
            Dec 2 '18 at 3:32










          • $begingroup$
            Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
            $endgroup$
            – Makina
            Dec 2 '18 at 3:38
















          1












          $begingroup$

          You already applied Ito's lemma to find
          $$
          e^{W(T)-T/2}=1+int_{0}^{T}e^{W(s)-s/2}dW(s).
          $$
          Next, note that
          begin{multline*}
          mathbb{E}left[int_{0}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]=int_{0}^{t}e^{W(s)-s/2}dW(s)+mathbb{E}left[int_{t}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]\=int_{0}^{t}e^{W(s)-s/2}dW(s).
          end{multline*}
          Therefore, applying conditional expectations to the first equation,
          $$
          e^{-T/2}M(t)=1+int_{0}^{t}e^{W(s)-s/2}dW(s).
          $$
          Multiplying by $e^{T/2}$ yields the desired result.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
            $endgroup$
            – Makina
            Nov 28 '18 at 6:41












          • $begingroup$
            I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
            $endgroup$
            – parsiad
            Dec 2 '18 at 3:32










          • $begingroup$
            Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
            $endgroup$
            – Makina
            Dec 2 '18 at 3:38














          1












          1








          1





          $begingroup$

          You already applied Ito's lemma to find
          $$
          e^{W(T)-T/2}=1+int_{0}^{T}e^{W(s)-s/2}dW(s).
          $$
          Next, note that
          begin{multline*}
          mathbb{E}left[int_{0}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]=int_{0}^{t}e^{W(s)-s/2}dW(s)+mathbb{E}left[int_{t}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]\=int_{0}^{t}e^{W(s)-s/2}dW(s).
          end{multline*}
          Therefore, applying conditional expectations to the first equation,
          $$
          e^{-T/2}M(t)=1+int_{0}^{t}e^{W(s)-s/2}dW(s).
          $$
          Multiplying by $e^{T/2}$ yields the desired result.






          share|cite|improve this answer









          $endgroup$



          You already applied Ito's lemma to find
          $$
          e^{W(T)-T/2}=1+int_{0}^{T}e^{W(s)-s/2}dW(s).
          $$
          Next, note that
          begin{multline*}
          mathbb{E}left[int_{0}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]=int_{0}^{t}e^{W(s)-s/2}dW(s)+mathbb{E}left[int_{t}^{T}e^{W(s)-s/2}dW(s)midmathcal{F}(t)right]\=int_{0}^{t}e^{W(s)-s/2}dW(s).
          end{multline*}
          Therefore, applying conditional expectations to the first equation,
          $$
          e^{-T/2}M(t)=1+int_{0}^{t}e^{W(s)-s/2}dW(s).
          $$
          Multiplying by $e^{T/2}$ yields the desired result.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 6:33









          parsiadparsiad

          17k32353




          17k32353












          • $begingroup$
            So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
            $endgroup$
            – Makina
            Nov 28 '18 at 6:41












          • $begingroup$
            I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
            $endgroup$
            – parsiad
            Dec 2 '18 at 3:32










          • $begingroup$
            Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
            $endgroup$
            – Makina
            Dec 2 '18 at 3:38


















          • $begingroup$
            So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
            $endgroup$
            – Makina
            Nov 28 '18 at 6:41












          • $begingroup$
            I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
            $endgroup$
            – parsiad
            Dec 2 '18 at 3:32










          • $begingroup$
            Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
            $endgroup$
            – Makina
            Dec 2 '18 at 3:38
















          $begingroup$
          So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
          $endgroup$
          – Makina
          Nov 28 '18 at 6:41






          $begingroup$
          So the answer will be $e^{frac{T - s}{2} + W(s)}$. Man, what a silly oversight by me ... thank you very much. May I ask if the adaptivity and integrability for the martingale property can be explained in a more sensible way? This MGF of an absolute value seems quite non-trivial
          $endgroup$
          – Makina
          Nov 28 '18 at 6:41














          $begingroup$
          I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
          $endgroup$
          – parsiad
          Dec 2 '18 at 3:32




          $begingroup$
          I think the MGF approach is fine (there might be some minor mistakes in the argument but the basic idea is fine). As for the martingale property, try proving it from first principles. If you are still having trouble, I can spend more time on this in the future.
          $endgroup$
          – parsiad
          Dec 2 '18 at 3:32












          $begingroup$
          Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
          $endgroup$
          – Makina
          Dec 2 '18 at 3:38




          $begingroup$
          Nah, I figured out it already. Thank you very much. The absolute value should be dropped since exponent is always positive. Removes all the complications.
          $endgroup$
          – Makina
          Dec 2 '18 at 3:38


















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