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
    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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 at 4:36









      Geoff Chen

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