I'm was reading an pdf about introduction to stochastic processes and the example say:
A coin it's tossed a few times. By each head the player win $1$ unit and by each tail the player loss $1$ unit.
So then in the example we have $n=6$ and $\omega = (h, h, t, t, t, t)$. This is
$X_1(\omega) = 1$
$X_2(\omega) = 2$
$X_3(\omega) = 1$
$X_4(\omega) = 0$
$X_5(\omega) = -1$
$X_6(\omega) = -2$
Next, the example fixs $t = 3$ for calcules the distribution of $X_3$ so the set of possibles states is:
So we have that:
$P{\{X_3 = -3\}} =\frac{1}{2^3} = \frac{1}8$
$P{\{X_3 = -1\}} =3\frac{1}{2^3} = \frac{3}8$
$P{\{X_3 = 1\}} =3\frac{1}{2^3} = \frac{3}8$
$P{\{X_3 = 3\}} =\frac{1}{2^3} = \frac{1}8$
Why $X_3 = -1$ and $X_3 = 1$ are multiplied by $3$?
