How to compute $mathbb{E}[X_{s}^{2}e^{lambda X_{s}}]$ where $(X_s)$ is a Brownian motion with drift $mu$?












3












$begingroup$


I'm working on a problem and at a certain point I ran into the problem as described in the title. We have that ${W_t,tgeq 0}$ is a Brownian motion and $mathscr{F}_t$ is the corresponding filtration. We have that $mu>0$ is given in the process ${X_t,tgeq 0}$ defined via $X_t:=mu t+W_t$.



I don't want to post the full problem I was solving yet, rather I'd like to know if what I ended up with is even solvable, because if not, I'll know I'm definitely wrong.



As posted in the title, I came at a point where I was left to compute the expectation:




$mathbb{E}[X_{s}^{2}e^{lambda X_{s}}]$




Earlier in the exercise (it consisted of multiple parts) I used the moment generating function for the normal distribution. However, as far as I know, I cannot take the $X_{s}^{2}$ out of the expectation, stopping me from applying the moment generating function.



Is this expectation solvable in a relatively easy way? If not, I'll know I'm wrong and start over.






probability-theory stochastic-processes brownian-motion expected-value







share|cite|improve this question











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  • 6




    $begingroup$
    Differentiate twice $E(e^{lambda X_t})$ with respect to $lambda$.
    $endgroup$
    – Did
    Jan 4 at 20:16










  • $begingroup$
    @Did That works. Thank you very much!
    $endgroup$
    – S. Crim
    Jan 5 at 8:50
















3












$begingroup$


I'm working on a problem and at a certain point I ran into the problem as described in the title. We have that ${W_t,tgeq 0}$ is a Brownian motion and $mathscr{F}_t$ is the corresponding filtration. We have that $mu>0$ is given in the process ${X_t,tgeq 0}$ defined via $X_t:=mu t+W_t$.



I don't want to post the full problem I was solving yet, rather I'd like to know if what I ended up with is even solvable, because if not, I'll know I'm definitely wrong.



As posted in the title, I came at a point where I was left to compute the expectation:




$mathbb{E}[X_{s}^{2}e^{lambda X_{s}}]$




Earlier in the exercise (it consisted of multiple parts) I used the moment generating function for the normal distribution. However, as far as I know, I cannot take the $X_{s}^{2}$ out of the expectation, stopping me from applying the moment generating function.



Is this expectation solvable in a relatively easy way? If not, I'll know I'm wrong and start over.






probability-theory stochastic-processes brownian-motion expected-value







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Differentiate twice $E(e^{lambda X_t})$ with respect to $lambda$.
    $endgroup$
    – Did
    Jan 4 at 20:16










  • $begingroup$
    @Did That works. Thank you very much!
    $endgroup$
    – S. Crim
    Jan 5 at 8:50














3












3








3





$begingroup$


I'm working on a problem and at a certain point I ran into the problem as described in the title. We have that ${W_t,tgeq 0}$ is a Brownian motion and $mathscr{F}_t$ is the corresponding filtration. We have that $mu>0$ is given in the process ${X_t,tgeq 0}$ defined via $X_t:=mu t+W_t$.



I don't want to post the full problem I was solving yet, rather I'd like to know if what I ended up with is even solvable, because if not, I'll know I'm definitely wrong.



As posted in the title, I came at a point where I was left to compute the expectation:




$mathbb{E}[X_{s}^{2}e^{lambda X_{s}}]$




Earlier in the exercise (it consisted of multiple parts) I used the moment generating function for the normal distribution. However, as far as I know, I cannot take the $X_{s}^{2}$ out of the expectation, stopping me from applying the moment generating function.



Is this expectation solvable in a relatively easy way? If not, I'll know I'm wrong and start over.






probability-theory stochastic-processes brownian-motion expected-value







share|cite|improve this question











$endgroup$




I'm working on a problem and at a certain point I ran into the problem as described in the title. We have that ${W_t,tgeq 0}$ is a Brownian motion and $mathscr{F}_t$ is the corresponding filtration. We have that $mu>0$ is given in the process ${X_t,tgeq 0}$ defined via $X_t:=mu t+W_t$.



I don't want to post the full problem I was solving yet, rather I'd like to know if what I ended up with is even solvable, because if not, I'll know I'm definitely wrong.



As posted in the title, I came at a point where I was left to compute the expectation:




$mathbb{E}[X_{s}^{2}e^{lambda X_{s}}]$




Earlier in the exercise (it consisted of multiple parts) I used the moment generating function for the normal distribution. However, as far as I know, I cannot take the $X_{s}^{2}$ out of the expectation, stopping me from applying the moment generating function.



Is this expectation solvable in a relatively easy way? If not, I'll know I'm wrong and start over.






probability-theory stochastic-processes brownian-motion expected-value




probability-theory stochastic-processes brownian-motion expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 14:16









Did

247k23223459




247k23223459










asked Jan 4 at 20:13









S. CrimS. Crim

36612




36612








  • 6




    $begingroup$
    Differentiate twice $E(e^{lambda X_t})$ with respect to $lambda$.
    $endgroup$
    – Did
    Jan 4 at 20:16










  • $begingroup$
    @Did That works. Thank you very much!
    $endgroup$
    – S. Crim
    Jan 5 at 8:50














  • 6




    $begingroup$
    Differentiate twice $E(e^{lambda X_t})$ with respect to $lambda$.
    $endgroup$
    – Did
    Jan 4 at 20:16










  • $begingroup$
    @Did That works. Thank you very much!
    $endgroup$
    – S. Crim
    Jan 5 at 8:50








6




6




$begingroup$
Differentiate twice $E(e^{lambda X_t})$ with respect to $lambda$.
$endgroup$
– Did
Jan 4 at 20:16




$begingroup$
Differentiate twice $E(e^{lambda X_t})$ with respect to $lambda$.
$endgroup$
– Did
Jan 4 at 20:16












$begingroup$
@Did That works. Thank you very much!
$endgroup$
– S. Crim
Jan 5 at 8:50




$begingroup$
@Did That works. Thank you very much!
$endgroup$
– S. Crim
Jan 5 at 8:50










1 Answer
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1












$begingroup$

Using the setting as stated in the question and working out the expectation by using the tip given in the comments, yields the following. The key observation is that the expectation we want to compute, equals the second derivative with respect to $lambda$ of the moment generating function for a normal distribution. The moment generating function for a $N(mu,sigma^2)$ is given by (with $X$ in this case a continuous random variable):



$mathbb{E}[e^{lambda X}]=e^{lambda mu + frac{1}{2}sigma^2 lambda^2}$



In our setting we have that $X_s = mu s +W_ssim N(mu s, s)$. So the moment generating function becomes:



$mathbb{E}[e^{lambda X_s}]=e^{lambda mu s + frac{1}{2}s lambda^2}$



Taking the first derivative with respect to $lambda$ yields:



$mathbb{E}[X_{s}e^{lambda X_s}]=(mu s+slambda)e^{lambda mu s + frac{1}{2}s lambda^2}$



Taking the derivative of the first derivative yields our desired answer:



$mathbb{E}[X_{s}^{2}e^{lambda X_s}]=(mu s+slambda)^{2}e^{lambda mu s + frac{1}{2}s lambda^2}+s e^{lambda mu s + frac{1}{2}s lambda^2}=((mu s+slambda)^{2}+s)e^{lambda mu s + frac{1}{2}s lambda^2} $






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    $begingroup$

    Using the setting as stated in the question and working out the expectation by using the tip given in the comments, yields the following. The key observation is that the expectation we want to compute, equals the second derivative with respect to $lambda$ of the moment generating function for a normal distribution. The moment generating function for a $N(mu,sigma^2)$ is given by (with $X$ in this case a continuous random variable):



    $mathbb{E}[e^{lambda X}]=e^{lambda mu + frac{1}{2}sigma^2 lambda^2}$



    In our setting we have that $X_s = mu s +W_ssim N(mu s, s)$. So the moment generating function becomes:



    $mathbb{E}[e^{lambda X_s}]=e^{lambda mu s + frac{1}{2}s lambda^2}$



    Taking the first derivative with respect to $lambda$ yields:



    $mathbb{E}[X_{s}e^{lambda X_s}]=(mu s+slambda)e^{lambda mu s + frac{1}{2}s lambda^2}$



    Taking the derivative of the first derivative yields our desired answer:



    $mathbb{E}[X_{s}^{2}e^{lambda X_s}]=(mu s+slambda)^{2}e^{lambda mu s + frac{1}{2}s lambda^2}+s e^{lambda mu s + frac{1}{2}s lambda^2}=((mu s+slambda)^{2}+s)e^{lambda mu s + frac{1}{2}s lambda^2} $






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Using the setting as stated in the question and working out the expectation by using the tip given in the comments, yields the following. The key observation is that the expectation we want to compute, equals the second derivative with respect to $lambda$ of the moment generating function for a normal distribution. The moment generating function for a $N(mu,sigma^2)$ is given by (with $X$ in this case a continuous random variable):



      $mathbb{E}[e^{lambda X}]=e^{lambda mu + frac{1}{2}sigma^2 lambda^2}$



      In our setting we have that $X_s = mu s +W_ssim N(mu s, s)$. So the moment generating function becomes:



      $mathbb{E}[e^{lambda X_s}]=e^{lambda mu s + frac{1}{2}s lambda^2}$



      Taking the first derivative with respect to $lambda$ yields:



      $mathbb{E}[X_{s}e^{lambda X_s}]=(mu s+slambda)e^{lambda mu s + frac{1}{2}s lambda^2}$



      Taking the derivative of the first derivative yields our desired answer:



      $mathbb{E}[X_{s}^{2}e^{lambda X_s}]=(mu s+slambda)^{2}e^{lambda mu s + frac{1}{2}s lambda^2}+s e^{lambda mu s + frac{1}{2}s lambda^2}=((mu s+slambda)^{2}+s)e^{lambda mu s + frac{1}{2}s lambda^2} $






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Using the setting as stated in the question and working out the expectation by using the tip given in the comments, yields the following. The key observation is that the expectation we want to compute, equals the second derivative with respect to $lambda$ of the moment generating function for a normal distribution. The moment generating function for a $N(mu,sigma^2)$ is given by (with $X$ in this case a continuous random variable):



        $mathbb{E}[e^{lambda X}]=e^{lambda mu + frac{1}{2}sigma^2 lambda^2}$



        In our setting we have that $X_s = mu s +W_ssim N(mu s, s)$. So the moment generating function becomes:



        $mathbb{E}[e^{lambda X_s}]=e^{lambda mu s + frac{1}{2}s lambda^2}$



        Taking the first derivative with respect to $lambda$ yields:



        $mathbb{E}[X_{s}e^{lambda X_s}]=(mu s+slambda)e^{lambda mu s + frac{1}{2}s lambda^2}$



        Taking the derivative of the first derivative yields our desired answer:



        $mathbb{E}[X_{s}^{2}e^{lambda X_s}]=(mu s+slambda)^{2}e^{lambda mu s + frac{1}{2}s lambda^2}+s e^{lambda mu s + frac{1}{2}s lambda^2}=((mu s+slambda)^{2}+s)e^{lambda mu s + frac{1}{2}s lambda^2} $






        share|cite|improve this answer









        $endgroup$



        Using the setting as stated in the question and working out the expectation by using the tip given in the comments, yields the following. The key observation is that the expectation we want to compute, equals the second derivative with respect to $lambda$ of the moment generating function for a normal distribution. The moment generating function for a $N(mu,sigma^2)$ is given by (with $X$ in this case a continuous random variable):



        $mathbb{E}[e^{lambda X}]=e^{lambda mu + frac{1}{2}sigma^2 lambda^2}$



        In our setting we have that $X_s = mu s +W_ssim N(mu s, s)$. So the moment generating function becomes:



        $mathbb{E}[e^{lambda X_s}]=e^{lambda mu s + frac{1}{2}s lambda^2}$



        Taking the first derivative with respect to $lambda$ yields:



        $mathbb{E}[X_{s}e^{lambda X_s}]=(mu s+slambda)e^{lambda mu s + frac{1}{2}s lambda^2}$



        Taking the derivative of the first derivative yields our desired answer:



        $mathbb{E}[X_{s}^{2}e^{lambda X_s}]=(mu s+slambda)^{2}e^{lambda mu s + frac{1}{2}s lambda^2}+s e^{lambda mu s + frac{1}{2}s lambda^2}=((mu s+slambda)^{2}+s)e^{lambda mu s + frac{1}{2}s lambda^2} $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 14:13









        S. CrimS. Crim

        36612




        36612






























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