Expectation in a stochastic differential equationWhat is Ito's lemma used for in quantitative finance?Question about the stochastic differential equation in the Merton modelComputation of ExpectationSquare of arithmetic brownian motion processBaxter & Rennie HJM: differentiating Ito integralSimple HJM model, differentiating the bond priceStochastic Leibniz ruleStochastic differential equation of a Brownian MotionHow to calculate the product of forward rates with different reset times using Ito's lemma?For an Ito Process, $dlnX neq fracdXX$ and $(dlnX)^2 = (fracdXX)^2$, but $dlnX neq pm fracdXX$

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Expectation in a stochastic differential equation


What is Ito's lemma used for in quantitative finance?Question about the stochastic differential equation in the Merton modelComputation of ExpectationSquare of arithmetic brownian motion processBaxter & Rennie HJM: differentiating Ito integralSimple HJM model, differentiating the bond priceStochastic Leibniz ruleStochastic differential equation of a Brownian MotionHow to calculate the product of forward rates with different reset times using Ito's lemma?For an Ito Process, $dlnX neq fracdXX$ and $(dlnX)^2 = (fracdXX)^2$, but $dlnX neq pm fracdXX$













2












$begingroup$


I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.



I used Ito's Lemma is arrive at the SDE:
beginalign
d(X_t) = frac12X_t dt + X_t dW_t
endalign

But how can I get the mean of $X_2$?










share|improve this question











$endgroup$
















    2












    $begingroup$


    I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.



    I used Ito's Lemma is arrive at the SDE:
    beginalign
    d(X_t) = frac12X_t dt + X_t dW_t
    endalign

    But how can I get the mean of $X_2$?










    share|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.



      I used Ito's Lemma is arrive at the SDE:
      beginalign
      d(X_t) = frac12X_t dt + X_t dW_t
      endalign

      But how can I get the mean of $X_2$?










      share|improve this question











      $endgroup$




      I'm new to stochastic calculus, I want to find the mean of $X_2$ with $X_t = exp(W_t)$, with $W_t$ a Wiener process.



      I used Ito's Lemma is arrive at the SDE:
      beginalign
      d(X_t) = frac12X_t dt + X_t dW_t
      endalign

      But how can I get the mean of $X_2$?







      itos-lemma sde






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 days ago







      Victor

















      asked 2 days ago









      VictorVictor

      764




      764




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Assuming you are talking about unconditional expectation, in general you have



          $$
          mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
          $$



          which yields



          $$
          mathbbE[X_t]= e^frac12 t
          $$



          Hence,



          $$ mathbbE[X_2]= e $$






          share|improve this answer








          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
            $endgroup$
            – Victor
            2 days ago






          • 1




            $begingroup$
            @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
            $endgroup$
            – RafaelC
            2 days ago











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Assuming you are talking about unconditional expectation, in general you have



          $$
          mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
          $$



          which yields



          $$
          mathbbE[X_t]= e^frac12 t
          $$



          Hence,



          $$ mathbbE[X_2]= e $$






          share|improve this answer








          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
            $endgroup$
            – Victor
            2 days ago






          • 1




            $begingroup$
            @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
            $endgroup$
            – RafaelC
            2 days ago















          3












          $begingroup$

          Assuming you are talking about unconditional expectation, in general you have



          $$
          mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
          $$



          which yields



          $$
          mathbbE[X_t]= e^frac12 t
          $$



          Hence,



          $$ mathbbE[X_2]= e $$






          share|improve this answer








          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
            $endgroup$
            – Victor
            2 days ago






          • 1




            $begingroup$
            @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
            $endgroup$
            – RafaelC
            2 days ago













          3












          3








          3





          $begingroup$

          Assuming you are talking about unconditional expectation, in general you have



          $$
          mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
          $$



          which yields



          $$
          mathbbE[X_t]= e^frac12 t
          $$



          Hence,



          $$ mathbbE[X_2]= e $$






          share|improve this answer








          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          Assuming you are talking about unconditional expectation, in general you have



          $$
          mathbbE[X_t] = mathbbE[e^W_t] = e^mathbbE[W_t] + frac12textVar(W_t)
          $$



          which yields



          $$
          mathbbE[X_t]= e^frac12 t
          $$



          Hence,



          $$ mathbbE[X_2]= e $$







          share|improve this answer








          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|improve this answer



          share|improve this answer






          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 2 days ago









          RafaelCRafaelC

          1463




          1463




          New contributor




          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          RafaelC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.











          • $begingroup$
            I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
            $endgroup$
            – Victor
            2 days ago






          • 1




            $begingroup$
            @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
            $endgroup$
            – RafaelC
            2 days ago
















          • $begingroup$
            I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
            $endgroup$
            – Victor
            2 days ago






          • 1




            $begingroup$
            @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
            $endgroup$
            – RafaelC
            2 days ago















          $begingroup$
          I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
          $endgroup$
          – Victor
          2 days ago




          $begingroup$
          I'm quite new to the theory. What is the name of the first equality and under which hypotheses is it true?
          $endgroup$
          – Victor
          2 days ago




          1




          1




          $begingroup$
          @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
          $endgroup$
          – RafaelC
          2 days ago




          $begingroup$
          @Victor the first equality comes from the moment-generating function of a normal. Take a look here for more details. In general, $mathbbE[e^X] = e^mu + frac12 sigma^2$ holds whenever $X sim mathcalN(mu, sigma^2)$
          $endgroup$
          – RafaelC
          2 days ago

















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