Discrete martingale property of $\tilde{X}_t=X_t-\sum_{r=0}^{t-1}\frac{X_T-X_r}{T-r}$

Question

This exercise is taken from the book "Stochastic Finance" of Föllmer and Schied. It is exercise 5.3.2.

Let $Y_1,\ldots,Y_T$ be iid random variables in $L^1$, which are almost surely not constant and satisfy $E[Y_k]=0$ for every $k=1,\ldots,T$.

Define $X_0:=1$ and for $t=1,\ldots, T$:

$$X_t:=X_0+\sum_{k=1}^tY_k$$

It is straightforward to show, that $X$ is a martingale with respect to the filtration $(\mathcal{F}_t)_{t=0,\ldots, T}$ with $\mathcal{F}_t=\sigma(Y_1,\ldots,Y_t)$ and $\mathcal{F}_0=\{\Omega,\emptyset\}$.

Now enlarge the filtration: $(\tilde{\mathcal{F}}_t)_{t=0,\ldots, T}$ with $\tilde{\mathcal{F}}_t=\sigma \big(\mathcal{F}_t,\sigma(X_T)\big)$.

I was able to show, that $X$ is no longer a martingale with respect to $(\tilde{\mathcal{F}}_t)_{t=0,\ldots, T}$, since for any $s<T$

$$E[X_T|\tilde{\mathcal{F}}_s]=X_T\ne X_s$$

Now I want to show that $(\tilde{X}_t)_{t=0,\ldots, T}$ is a martingale with respect to $(\tilde{\mathcal{F}}_t)_{t=0,\ldots, T}$, whereas

$$\tilde{X}_t:=X_t-\sum_{r=0}^{t-1}\frac{X_T-X_r}{T-r}$$

Since $X$ is a martingale one can conclude that $E[|\tilde{X}_t|]<\infty$.

$\tilde{X}_t$ is an continuous image of $X_0,\ldots,X_{t-1},X_T$, therefore $\tilde{\mathcal{F}}_t$-measurable.

Let $s<t$.

$$\begin{align} E[\tilde{X}_t|\tilde{\mathcal{F}}_s] &=E\Big[X_t-\sum_{r=0}^{t-1}\frac{X_T-X_r}{T-r}\Big|\tilde{\mathcal{F}}_s\Big]\\ &=\tilde{X}_s+E\Big[(X_t-X_s)-\sum_{r=s}^{t-1}\frac{X_T-X_r}{T-r}\Big|\tilde{\mathcal{F}}_s\Big] \end{align}$$

But I fail to show that $$E\Big[(X_t-X_s)-\sum_{r=s}^{t-1}\frac{X_T-X_r}{T-r}\Big|\tilde{\mathcal{F}}_s\Big]=0$$

I did following $$\begin{align}X_t-X_s&=\sum_{k=s+1}^t Y_k=(t-s)\frac{1}{t-s}\sum_{k=s+1}^t Y_k=(t-s)E\Big[\text{ }Y_t\text{ }\Big|\sum_{k=s+1}^t Y_k\Big]\end{align}$$

$$\begin{align}\frac{1}{T-r}(X_T-X_r)&=\frac{1}{T-r}\sum_{k=r+1}^TY_k=E\Big[\text{ }Y_t\text{ }\Big|\sum_{k=r+1}^TY_k\Big]\end{align}$$

which gives

$$(X_t-X_s)-\sum_{r=s}^{t-1}\frac{X_T-X_r}{T-r}=\sum_{r=s}^{t-1}\bigg(E\Big[\text{ }Y_t\text{ }\Big|\sum_{k=s+1}^t Y_k\Big]-E\Big[\text{ }Y_t\text{ }\Big|\sum_{k=r+1}^TY_k\Big]\bigg)$$

and almost looks like $0$, but I am not quite convinced yet, why this is $0$ after taking the conditional expectation with respect to $\tilde{\mathcal{F}}_s$.

I'd appreciate some help on this. Thanks for your attention!

Answer 1

This was a fun problem. The trick is to understand how knowing $X_T$ changes the conditional expectation. We can very easily simplify $\mathbb{E}[\tilde{X}_t|\tilde{\mathcal{F}}_{t-1}]$:

$$\mathbb{E}[\tilde{X}_t|\tilde{\mathcal{F}}_{t-1}] = X_{t-1} - \sum_{r=0}^{t-1}\frac{X_T - X_r}{T-r} + \mathbb{E}[Y_t|\tilde{\mathcal{F}}_{t-1}].$$

So how can we compute $\mathbb{E}[Y_t|\tilde{\mathcal{F}}_{t-1}]$? Notice a few things. First, even conditioned on $\tilde{F}_{t-1}$, $Y_t,\dots,Y_T$ are identically distributed (although they aren't independent anymore). Furthermore, we know that,

$$X_{t-1} + \sum_{r=t}^T Y_r = X_T.$$

Subtracting $X_{t-1}$ from both sides, taking the conditional expectation and applying linearity of the conditional expectation yields,

$$(T-t+1)\mathbb{E}[Y_t|\tilde{\mathcal{F}}_{t-1}] = \mathbb{E}\left[\sum_{r=t}^T Y_r\middle|\tilde{\mathcal{F}}_{t-1}\right] = \mathbb{E}[X_T - X_{t-1}|\tilde{\mathcal{F}}_{t-1}] = X_T - X_{t-1}.$$

So,

$$\mathbb{E}[Y_t|\tilde{\mathcal{F}}_{t-1}] = \frac{X_T - X_{t-1}}{T-t+1}.$$

Which gives us,

$$\mathbb{E}[\tilde{X_t}|\tilde{\mathcal{F}}_{t-1}] = X_{t-1} - \sum_{r=0}^{t-1}\frac{X_T - X_r}{T-r} + \frac{X_T - X_{t-1}}{T-t+1} = X_{t-1} + \sum_{r=0}^{t-2}\frac{X_T - X_r}{T-r} = \tilde{X}_{t-1}.$$