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.
exponential-function brownian-motion martingales expected-value filtrations
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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.
exponential-function brownian-motion martingales expected-value filtrations
$endgroup$
add a comment |
$begingroup$
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.
exponential-function brownian-motion martingales expected-value filtrations
$endgroup$
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
exponential-function brownian-motion martingales expected-value filtrations
edited Mar 18 at 14:06
rs4rs35
asked Mar 18 at 13:53
rs4rs35rs4rs35
447
447
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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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1 Answer
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1 Answer
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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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Mar 18 at 16:54
John DawkinsJohn Dawkins
13.3k11017
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