Why is the ratio of two returns martingale?Futures pricing and futures price process under the real world measureEvaluating Expectation of stochastic processGirsanov theorem for Ito diffusion processWhy is a discounted price process a local martingale under the Risk Neutral Measure?Girsanov theorem calculations helpHelp with change of measure and martingalesExpectation $ mathbb E^mathbb P^tau_tleft[ X_tau(T)Y_tauright] $ and measure change.How to solve this change in numerairestochastic representation using Girsanov's formulaWhat is the Radon-Nikodym density $dP^∗ /d P$ of the unique $P^ ∗ ∼ P$ such that the discounted price $S^*_t := S_t /B_t$ is a $P^∗$-martingale
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Why is the ratio of two returns martingale?
Futures pricing and futures price process under the real world measureEvaluating Expectation of stochastic processGirsanov theorem for Ito diffusion processWhy is a discounted price process a local martingale under the Risk Neutral Measure?Girsanov theorem calculations helpHelp with change of measure and martingalesExpectation $ mathbb E^mathbb P^tau_tleft[ X_tau(T)Y_tauright] $ and measure change.How to solve this change in numerairestochastic representation using Girsanov's formulaWhat is the Radon-Nikodym density $dP^∗ /d P$ of the unique $P^ ∗ ∼ P$ such that the discounted price $S^*_t := S_t /B_t$ is a $P^∗$-martingale
$begingroup$
If we have a ratio of two fund returns, under the risk-neutral measure $y=frac left. F_T^2 middle/ F_0^2 right.left. F_T^1 middle/ F_0^1 right.$.
Let $X_t$ represents the price of asset 2 at time $t$.
$$Xt=E_tleft[fracF_T^2expleft(int_0^T r_u d_u right)right]$$
Then it can be seen that
$$X_t=fracF_t^2 expleft( int_0^t r_u d_u right ) underbrace E_tleft[ frac left. F_T^2 middle/ F_t^2 right. expleft(int_t^T r_u d_u right)right] _=1$$
My question: why is the last expectation equal to $1$?
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-integrals
$endgroup$
add a comment |
$begingroup$
If we have a ratio of two fund returns, under the risk-neutral measure $y=frac left. F_T^2 middle/ F_0^2 right.left. F_T^1 middle/ F_0^1 right.$.
Let $X_t$ represents the price of asset 2 at time $t$.
$$Xt=E_tleft[fracF_T^2expleft(int_0^T r_u d_u right)right]$$
Then it can be seen that
$$X_t=fracF_t^2 expleft( int_0^t r_u d_u right ) underbrace E_tleft[ frac left. F_T^2 middle/ F_t^2 right. expleft(int_t^T r_u d_u right)right] _=1$$
My question: why is the last expectation equal to $1$?
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-integrals
$endgroup$
$begingroup$
Please check if I messed up the math in my edit. BTW if you're doing this kind of big fractions a lot you should learn to use the middle slash, as well as a good habit of using the automatically sizing delimiters.
$endgroup$
– Lee David Chung Lin
2 days ago
$begingroup$
Thanks for your help. I appreciate it and will follow your advice in my future post
$endgroup$
– veryBigman
yesterday
add a comment |
$begingroup$
If we have a ratio of two fund returns, under the risk-neutral measure $y=frac left. F_T^2 middle/ F_0^2 right.left. F_T^1 middle/ F_0^1 right.$.
Let $X_t$ represents the price of asset 2 at time $t$.
$$Xt=E_tleft[fracF_T^2expleft(int_0^T r_u d_u right)right]$$
Then it can be seen that
$$X_t=fracF_t^2 expleft( int_0^t r_u d_u right ) underbrace E_tleft[ frac left. F_T^2 middle/ F_t^2 right. expleft(int_t^T r_u d_u right)right] _=1$$
My question: why is the last expectation equal to $1$?
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-integrals
$endgroup$
If we have a ratio of two fund returns, under the risk-neutral measure $y=frac left. F_T^2 middle/ F_0^2 right.left. F_T^1 middle/ F_0^1 right.$.
Let $X_t$ represents the price of asset 2 at time $t$.
$$Xt=E_tleft[fracF_T^2expleft(int_0^T r_u d_u right)right]$$
Then it can be seen that
$$X_t=fracF_t^2 expleft( int_0^t r_u d_u right ) underbrace E_tleft[ frac left. F_T^2 middle/ F_t^2 right. expleft(int_t^T r_u d_u right)right] _=1$$
My question: why is the last expectation equal to $1$?
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-integrals
probability-theory stochastic-processes stochastic-calculus conditional-probability stochastic-integrals
edited 2 days ago
Lee David Chung Lin
4,38031242
4,38031242
asked 2 days ago
veryBigmanveryBigman
153
153
$begingroup$
Please check if I messed up the math in my edit. BTW if you're doing this kind of big fractions a lot you should learn to use the middle slash, as well as a good habit of using the automatically sizing delimiters.
$endgroup$
– Lee David Chung Lin
2 days ago
$begingroup$
Thanks for your help. I appreciate it and will follow your advice in my future post
$endgroup$
– veryBigman
yesterday
add a comment |
$begingroup$
Please check if I messed up the math in my edit. BTW if you're doing this kind of big fractions a lot you should learn to use the middle slash, as well as a good habit of using the automatically sizing delimiters.
$endgroup$
– Lee David Chung Lin
2 days ago
$begingroup$
Thanks for your help. I appreciate it and will follow your advice in my future post
$endgroup$
– veryBigman
yesterday
$begingroup$
Please check if I messed up the math in my edit. BTW if you're doing this kind of big fractions a lot you should learn to use the middle slash, as well as a good habit of using the automatically sizing delimiters.
$endgroup$
– Lee David Chung Lin
2 days ago
$begingroup$
Please check if I messed up the math in my edit. BTW if you're doing this kind of big fractions a lot you should learn to use the middle slash, as well as a good habit of using the automatically sizing delimiters.
$endgroup$
– Lee David Chung Lin
2 days ago
$begingroup$
Thanks for your help. I appreciate it and will follow your advice in my future post
$endgroup$
– veryBigman
yesterday
$begingroup$
Thanks for your help. I appreciate it and will follow your advice in my future post
$endgroup$
– veryBigman
yesterday
add a comment |
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$begingroup$
Please check if I messed up the math in my edit. BTW if you're doing this kind of big fractions a lot you should learn to use the middle slash, as well as a good habit of using the automatically sizing delimiters.
$endgroup$
– Lee David Chung Lin
2 days ago
$begingroup$
Thanks for your help. I appreciate it and will follow your advice in my future post
$endgroup$
– veryBigman
yesterday