Given a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_n\}, \mathbb P)$, let $X = (X_n)_{n \in \mathbb N}$ and $Y = (Y_n)_{n \in \mathbb N}$ be $(\{\mathscr F_n\}, \mathbb P)-$martingales and $T$ be a $\{\mathscr F_n\}$-stopping time s.t. $X_T = Y_T$

Prove $Z_n = X_n1_{n \le T} + Y_n1_{n-1\ge T}$ is a $(\{\mathscr F_n\}, \mathbb P)-$martingale.

What I tried:

  1. Integrable:

$X_n, Y_n, 1_{n \le T}, 1_{n-1\ge T}$ are integrable

  1. Adapted to $\{\mathscr F_n\}$:

$X_n, Y_n, 1_{n \le T}, 1_{n-1\ge T}$ are adapted because $X$ and $Y$ are martingales and $T$ is a stopping time

  1. $$E[Z_n | \mathscr F_{m}] = Z_{m}$$

Not sure I can do that directly, but I guess can make use of:

$$E[Z_n | \mathscr F_{m}] = E[E[Z_n|\mathscr F_{n-1}] | \mathscr F_{m}] = E[Z_{n-1} | \mathscr F_{m}] = ... = E[Z_m | \mathscr F_{m}] = Z_m$$

by showing that $Z_{n-1} = E[Z_n|\mathscr F_{n-1}]$:

$$E[Z_n | \mathscr F_{n-1}] = E[X_n1_{n \le T} + Y_n1_{n-1\ge T} | \mathscr F_{n-1}]$$

$$= E[X_n1_{n \le T} | \mathscr F_{n-1}] + E[Y_n1_{n-1\ge T} | \mathscr F_{n-1}]$$

$$= 1_{n \le T} E[X_n | \mathscr F_{n-1}] + 1_{n-1\ge T}E[Y_n | \mathscr F_{n-1}]$$

$$= 1_{n \le T} X_{n-1} + 1_{n-1\ge T}Y_{n-1}$$

$\because Y_{n-1}1_{T=n-1} = Y_{T}1_{T=n-1} = X_{T}1_{T=n-1} = X_{n-1}1_{T=n-1}$, we have

$$= 1_{(n-1) \le T} X_{n-1} + 1_{n-2\ge T}Y_{n-1}$$

Is that right?

  • 1
    $\begingroup$ Is there any relationship between the two martingales? I don't think this is true as currently stated. For example take $T$ constant, and let $X_n=1$ for all $n$, $Y_n=2$ for all $n$. $\endgroup$ Dec 14, 2015 at 22:06

1 Answer 1


Your argument is basically correct.

Two comments, points where you could make your argument clearer or make your job easier :

  1. Integrability: the product of two integrable random variables is not necessarily integrable, however in this case $1_{n \le T}$ and $1_{n-1\ge T}$ are bounded (by 1) so the products $X_n1_{n \le T}$ and $Y_n1_{n-1\ge T}$ are integrable.
  2. Martingale property. This would often be stated as $Z_{n-1} = E[Z_n|\mathscr F_{n-1}]$ (as in Williams' book "Probability with Martingales", Section 10.3), so you don't need to start from the more general property $E[Z_n | \mathscr F_{m}] = Z_{m}$ (for all $n>m$).
  • $\begingroup$ Thanks S. Catterall. ^-^ Integrability: I knew that. Apologies for not making clear. 2 Ah, forgot about that. Guess it depends on the text. $\endgroup$
    – BCLC
    Dec 15, 2015 at 20:09
  • 1
    $\begingroup$ Good point. Maybe there are other texts that use the more general martingale property definition... but Williams' definition makes your work slightly easier. $\endgroup$ Dec 15, 2015 at 20:18

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