Prove $|M_{T \wedge n}| \le c + K$

97 Views Asked by At

From Williams' Probability with Martingales


enter image description here


enter image description here


  1. Is $\sigma_k^2$ random (and not constant)? How can that be? As far as I know unconditional variance and unconditional expectations are supposed to be constant.

  1. How do we know $|M_n^T| \le c + K$?

This is what I tried:

$$M_n^T = M_{T \wedge n} = M_0 + \sum_{k=1}^{T \wedge n} (M_k - M_{k-1})$$

$$\to |M_n^T| \le |M_0| + \sum_{k=1}^{T \wedge n} |M_k - M_{k-1}|$$

$$\le |M_0| + \sum_{k=1}^{T \wedge n} |X_k|$$

$$\le c + \sum_{k=1}^{T \wedge n} |X_k|$$

$$\le c + \sum_{k=1}^{T \wedge n} K$$

$$\le c + (T \wedge n - 1 + 1)(K)$$

$$ = c + (T \wedge n)(K)$$

I'm stuck. How can I approach this?