Random Process at a particular time instance

Question

I was studying Random Process and I thought I understood what it was all about until I came across this example. Consider a random experiment of tossing a coin with sample space S = {H, T} The sample functions are given by:

$X(t, H) = x_1(t) = \sin(\omega_1t) \tag{1}$

$X(t, T) = x_2(t) = \sin(\omega_2t) \tag{2}$

Now, I understand that random process at any particular time instance say $t_k$ gives me a random variable. Also, the sum of probabilities of a random variable should be 1 i.e. $$X(t_k,H) + X(t_k,T) = 1 \tag{3}$$ subsequently in general no matter where I sample the random process I should have, $$\sin(\omega_1t) + \sin(\omega_2t) = 1 \tag{4}$$

But the Equation 4 won't be true for all values of t.

Question 1. Where am I going wrong above?

Question 2. If I take the value of $X(t,\lambda)$ at a particular time $t_k$and at a particular $\lambda_i$ the value $X(t_k,\lambda_i)$ will be a number. This number is equal to the probability of $X(t,\lambda)$ taking value $\lambda_i$ at time instance $t_k$. Am I correct?

Answer 1

You misunderstand what's going on here. The random process $X(t)$ is of the form

$$X(t)=\sin(\Omega t)\tag{1}$$

where $\Omega$ is a random variable with PDF

$$p_{\Omega}(\omega)=P[H]\cdot \delta(\omega-\omega_1)+ P[T]\cdot \delta(\omega-\omega_2)\tag{2}$$

where $P[H]$ and $P[T]$ are the probabilities of heads and tails, respectively. Of course we must have $P[H]+P[T]=1$. According to $(2)$, the RV $\Omega$ can only take on two values, either $\omega_1$ (in the case of heads) or $\omega_2$ (in the case of tails). Because of that, there are only two possible realizations of $X(t)$: either $\sin(\omega_1t)$ or $\sin(\omega_2t)$. Note that the value of a realization of a random process at a certain time has nothing to do with probability.

Answer 2

Your equation (3) and (4) are wrong.

When you say

Also, the sum of probabilities of a random variable should be 1

you actually mean the total sum of the values of probability mass function $P_{X_k}[x]$ equals one; i,e.

$$ \sum_{k} P_X[x_k] = 1 $$

(or the integral of the probability density funtion $f_{X_k}(x)$ equals one as $ \int_{-\infty}^{\infty} f_X(x) dx = 1 $ )

But in your equation (3) and (4), you are not summing the values of probability mass function but the values of the sample function itself, which leads to te wrong conclusion and confussion.

To reach the correct conclusion, in your example RP, the pdf associated with each RV $X_{t_k}$ is $$ f_{X_k}(x) = P(H) \delta( x- \sin( \omega_1 t_k) ) + P(T) \delta(x- \sin( \omega_2 t_k)) $$

which integrastes (sums) to one as

$$ \begin{align} \int f_X(x) dx &= \int \left( P(H) \delta(x- \sin( \omega_1 t_k)) + P(T) \delta(z- \sin( \omega_2 t_k)) \right) dx \\ &= P(H) \int \delta(x- \sin( \omega_1 t_k)) dx + P(T) \int \delta(x- \sin( \omega_2 t_k)) dx \\ &= P(H)+P(T) = 1 \end{align} $$