Let $Z=(Z_1,Z_2,Z_3)$ be a triple coin toss with $p=1/4$. Let $Y:=\frac{1}{2}Z_1 + \left(\frac{1}{2}\right)^2Z_2+\left(\frac{1}{2}\right)^3Z_3$ be a random variable $\in [0,1]$. Calculate the value F (b) for the distribution function of Y for
i) b = $\frac{1}{2}$ ii)b = $\frac{3}{8}$ iii) b = $\frac{9}{16}$ iv)b = 2
Sketch the function F. Does the distribution of Y has a density?
Approach: The distribution function is defined as the following:
F (b) := $F_x$ (b) = P (X $\leq$ b), b $\in$ R.
The density is
F (x) = $$\int_{-∞}^{x} f(a) da$$ x $\in$ R
$$\int_{-∞}^{x} da = 1 $$
I don't know how to solve this task with these formulas or even if the task should be solved with this? It's the only thing our script says about distribution functions and density. :/
Since this problem only has 8 outcomes, you might find it easiest and more intuitive to compute the distribution function by generating each outcome and computing the value of the random variable Y. The 8 outcomes for $(Z_1, Z_2, Z_3)$ are:
(0,0,0)
(0,0,1)
(0,1,0)
(0,1,1)
(1,0,0)
(1,0,1)
(1,1,0)
(1,1,1)
Each of these corresponds to a different value for Y, and the probability of that value will be the product of the individual probabilities for $(Z_1, Z_2, Z_3)$. For example, for (0,0,1), the probability will be 3/4*3/4*1/4. I have assumed that "heads" = 1, "tails" =0, and P("heads")= 1/4, if I understand/infer the problem definition correctly. Hopefully, this will get you going in the right direction. I hope this helps.