# How to find the Cumulative Distributive Function of a element in a Stochastic Process?

Let $$X$$ a random variable with distribution Bernoulli with parameter $$p$$. For every $$t\geq0$$ the variable $$X_t$$ is defined as:

\begin{align} X_t=\begin{cases} cos(\pi t) & X=0 \\sin(\pi t) & X=1\end{cases} \end{align}

Find the CDF of $$X_t$$ and $$\mathbb{E}(X_t)$$

If i use the definition of CDF then (by the law of total probability):

$$P(X_t\leq x)=P(cos(\pi t)\leq x)\cdot (1-p)+P(sin(\pi t)\leq x)\cdot p$$

But, i'm stuck here.

1. It is not assumed that when $$x$$ tends to infinity then the CDF tends to $$1$$?
2. Maybe $$X_t$$ is distributed similar to a Bernoulli?
• Note that $cos,sin$ are bounded by $1$, so for $x \geq 1$, $P(cos(\pi t)≤x) = P(sin(pi t)≤x) = 1$, and so $P(X_t \leq x) = 1-p + p = 1$. Now it looks to be as if this is a discrete process where $X_t$ is independent of $X_{t-k}$. If it is indeed discrete, then $X_t$ takes values $-1,0,1$ only. Consider $cos((2k+1)\pi) = -1$, and so $X_t$ is not exactly bernoulli, but may be dependent slightly on the parity of $t$. – blanchey Apr 7 at 20:08
• @blanchey The exercise says that it is a continuous parameter. – retro_var Apr 7 at 22:25