Discounted stock is a martingale?
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I'm currently confused with the discount martingale.
From my class, I have been told the discounted stock is a martingale.
I know the fact under risk neutral, we have
$$E(e^{-r(T-t)}S_T)=S_t$$
So we say $e^{-r(T-t)}S_T$ is a martingale.
However, if we move $e^{rt}$ part to the right hand side, we have the following
$$E(e^{-rT}S_T)=e^{-rt}S_t$$
So should we say $e^{-rT}S_T$ is a martingale. But why people say $e^{-r(T-t)}S_T$ is a martingale?
martingales
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up vote
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I'm currently confused with the discount martingale.
From my class, I have been told the discounted stock is a martingale.
I know the fact under risk neutral, we have
$$E(e^{-r(T-t)}S_T)=S_t$$
So we say $e^{-r(T-t)}S_T$ is a martingale.
However, if we move $e^{rt}$ part to the right hand side, we have the following
$$E(e^{-rT}S_T)=e^{-rt}S_t$$
So should we say $e^{-rT}S_T$ is a martingale. But why people say $e^{-r(T-t)}S_T$ is a martingale?
martingales
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm currently confused with the discount martingale.
From my class, I have been told the discounted stock is a martingale.
I know the fact under risk neutral, we have
$$E(e^{-r(T-t)}S_T)=S_t$$
So we say $e^{-r(T-t)}S_T$ is a martingale.
However, if we move $e^{rt}$ part to the right hand side, we have the following
$$E(e^{-rT}S_T)=e^{-rt}S_t$$
So should we say $e^{-rT}S_T$ is a martingale. But why people say $e^{-r(T-t)}S_T$ is a martingale?
martingales
I'm currently confused with the discount martingale.
From my class, I have been told the discounted stock is a martingale.
I know the fact under risk neutral, we have
$$E(e^{-r(T-t)}S_T)=S_t$$
So we say $e^{-r(T-t)}S_T$ is a martingale.
However, if we move $e^{rt}$ part to the right hand side, we have the following
$$E(e^{-rT}S_T)=e^{-rt}S_t$$
So should we say $e^{-rT}S_T$ is a martingale. But why people say $e^{-r(T-t)}S_T$ is a martingale?
martingales
martingales
asked Nov 29 at 4:36
Geoff Chen
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