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?
  • 1
    $\begingroup$ 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 $t$. $\endgroup$ – blanchey Apr 7 at 20:08
  • $\begingroup$ @blanchey The exercise says that it is a continuous parameter. $\endgroup$ – retro_var Apr 7 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.