Find values of $a$ and $lambda$ for which $Z_{0}e^{at+bW_{t}}-lambda t$ is a martingaleFor what value of...

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Find values of $a$ and $lambda$ for which $Z_{0}e^{at+bW_{t}}-lambda t$ is a martingale


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Find values of $a$ and $lambda$ for which $Z(t)=Z_{0}e^{at+bW_{t}}-lambda t$ is a martingale. In here $W_{t}$ is a Brownian motion and $a,binmathbb{R}$ can be positive as well as negative, since $b$ is derived by substracting one variance from another.



I cannot find a form for which this is a martingale except the form of the exponential martingale $$Ccdot e^{alpha W_{t}-frac{alpha^{2}t}{2}}$$ with $C$ a constant. This would imply that only for the values $$a=frac{-b^{2}}{2}$$ and $$lambda=0$$ $Z(t)$ is a martingale. Is this correct or is there another form for which $Z(t)$ is a martingle, since I have to find a solution for $lambdainmathbb{R}^{+}$, but this is quite an ambiguous expression.










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    $begingroup$


    Find values of $a$ and $lambda$ for which $Z(t)=Z_{0}e^{at+bW_{t}}-lambda t$ is a martingale. In here $W_{t}$ is a Brownian motion and $a,binmathbb{R}$ can be positive as well as negative, since $b$ is derived by substracting one variance from another.



    I cannot find a form for which this is a martingale except the form of the exponential martingale $$Ccdot e^{alpha W_{t}-frac{alpha^{2}t}{2}}$$ with $C$ a constant. This would imply that only for the values $$a=frac{-b^{2}}{2}$$ and $$lambda=0$$ $Z(t)$ is a martingale. Is this correct or is there another form for which $Z(t)$ is a martingle, since I have to find a solution for $lambdainmathbb{R}^{+}$, but this is quite an ambiguous expression.










    share|cite|improve this question











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      0





      $begingroup$


      Find values of $a$ and $lambda$ for which $Z(t)=Z_{0}e^{at+bW_{t}}-lambda t$ is a martingale. In here $W_{t}$ is a Brownian motion and $a,binmathbb{R}$ can be positive as well as negative, since $b$ is derived by substracting one variance from another.



      I cannot find a form for which this is a martingale except the form of the exponential martingale $$Ccdot e^{alpha W_{t}-frac{alpha^{2}t}{2}}$$ with $C$ a constant. This would imply that only for the values $$a=frac{-b^{2}}{2}$$ and $$lambda=0$$ $Z(t)$ is a martingale. Is this correct or is there another form for which $Z(t)$ is a martingle, since I have to find a solution for $lambdainmathbb{R}^{+}$, but this is quite an ambiguous expression.










      share|cite|improve this question











      $endgroup$




      Find values of $a$ and $lambda$ for which $Z(t)=Z_{0}e^{at+bW_{t}}-lambda t$ is a martingale. In here $W_{t}$ is a Brownian motion and $a,binmathbb{R}$ can be positive as well as negative, since $b$ is derived by substracting one variance from another.



      I cannot find a form for which this is a martingale except the form of the exponential martingale $$Ccdot e^{alpha W_{t}-frac{alpha^{2}t}{2}}$$ with $C$ a constant. This would imply that only for the values $$a=frac{-b^{2}}{2}$$ and $$lambda=0$$ $Z(t)$ is a martingale. Is this correct or is there another form for which $Z(t)$ is a martingle, since I have to find a solution for $lambdainmathbb{R}^{+}$, but this is quite an ambiguous expression.







      exponential-function brownian-motion martingales expected-value filtrations






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      edited Mar 18 at 14:06







      rs4rs35

















      asked Mar 18 at 13:53









      rs4rs35rs4rs35

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          $begingroup$

          The mean of $Z(t)$ is $E[Z_0]cdot e^{at+b^2t/2}-lambda t$, which is constant (in $t$) if and only if $a+b^2/2=0=lambda$. This means that the sufficient condition you have found is also necessary, for $Z(t)$ to be a martingale.






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            $begingroup$

            The mean of $Z(t)$ is $E[Z_0]cdot e^{at+b^2t/2}-lambda t$, which is constant (in $t$) if and only if $a+b^2/2=0=lambda$. This means that the sufficient condition you have found is also necessary, for $Z(t)$ to be a martingale.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The mean of $Z(t)$ is $E[Z_0]cdot e^{at+b^2t/2}-lambda t$, which is constant (in $t$) if and only if $a+b^2/2=0=lambda$. This means that the sufficient condition you have found is also necessary, for $Z(t)$ to be a martingale.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The mean of $Z(t)$ is $E[Z_0]cdot e^{at+b^2t/2}-lambda t$, which is constant (in $t$) if and only if $a+b^2/2=0=lambda$. This means that the sufficient condition you have found is also necessary, for $Z(t)$ to be a martingale.






                share|cite|improve this answer









                $endgroup$



                The mean of $Z(t)$ is $E[Z_0]cdot e^{at+b^2t/2}-lambda t$, which is constant (in $t$) if and only if $a+b^2/2=0=lambda$. This means that the sufficient condition you have found is also necessary, for $Z(t)$ to be a martingale.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 18 at 16:54









                John DawkinsJohn Dawkins

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                13.3k11017






























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