Finding the distribution of a piecewise defined variable

Question

Let $X$ be a Bernoulli random variable with parameter $p$, $X_t$ be defined as follows: $$X_t =\begin{cases} \cos(\pi \cdot t) &\text{if}\quad X=0 \\ \sin(\pi \cdot t) &\text{if}\quad X=1 \end {cases} $$

Is the probability distribution of $X_t$ given by $f(t) = p\cdot \cos(\pi \cdot t) + (1-p) \cdot \sin(\pi \cdot t)$?

It looks like it but I'm not really sure, any help and explanation would be great, thank you. Also with it how can I calculate it's expected value, do I just integrate $x\cdot f$ over $\mathbb{R}$?

Answer 1

The $t$ in $X_t$ is not a density variable but the index for a sequence of random variables, so for each $t$, $X_t$ is a random variable.

For example: $X_1$ is $\cos(\pi)=-1$ if $X=0$ (Probability $1-p$) and $X_1$ is $\sin(\pi)=0$ if $X=1$ (Probability $p$).

The distribution function for every $X_t$ is discrete, for $X_1$ lets denote the distribution function as $F_1$, then it holds:

$F_1(x)=P(X_1\leq x)=0$ for all $x <-1$

$F_1(x)=1-p$ for all $-1\leq x<0$

$F_1(x)=1$ for $0\leq x$

Edit: The expected value for a discrete Random variable is $\sum xP(X=x)$ over the discrete set of $x$ with $P(X=x)>0$

$E[X_1]=(-1)(1-p)+0p=p-1$