# Stochastic Process | Doubt with an example

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 $$3$$?

• Welcome to Math.SE! Please use MathJax for the top part of your question. For some basic information about writing math at this site see e.g. basic help on MathJax notation, MathJax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Dec 15, 2020 at 12:39
• There's a typo in $X_6(\omega)$. Its value should be $-2$ as $X_5(\omega) = -1$, and you state that one loses $1$ unit in case of tail. Dec 15, 2020 at 12:41
• You see that each of $1$'s and $-1$'s appears times in your tree diagram. Each occurrence accounts for $\frac{1}{8}$ probability, so the probability is $\frac{3}{8}$. Alternatively, note that in order to have $X_3=1$. So you must have two H's and one T: $$\mathbb{P}(X_3=1)=\mathbb{P}(\text{HHT})+\mathbb{P}(\text{HTH})+\mathbb{P}(\text{THH})=\frac{3}{8}.$$ This idea generalizes and can be dealt with binomial distribution. Dec 15, 2020 at 12:42