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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. :/

J. W. Tanner
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Marceline
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1 Answers1

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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.

ad2004
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  • Thanks for you reply, ad2004. I already calculated the distribution weights in the task before. Since we have 8 possib le outcomes we have one time the possibility $\frac{27}{64}$ (for 0,0,0), three times $\frac{9}{64}$ and $\frac{3}{64}$ and one time $\frac{1}{64}$, which is sumed up = 1. That are the distribution weights. But how can I get the distribution function from that? – Marceline Nov 27 '19 at 23:31
  • Since you are looking for the distribution function for Y, you need to compute the value of Y for each of those outcomes. For example, for (0,0,0), Y= 1/2(0) + 1/4(0) + 1/8(0) = 0, with probability 27/64. For (0,0,1), Y = 1/2(0)+1/4(0)+1/8(1) = 1/8 with probability 9/64, etc... If you do this for all outcomes, you can determine F(b). For example, you might have F(b=1/2). From the definition, you can see that this is $ P(Y \le 1/2)$, and you can determine this from the sum of the probabilities of each Y value $\le 1/2$. I hope this helps. – ad2004 Nov 27 '19 at 23:58
  • Thank you, ad2004. To me, it's not clear which probabilities should be sumed up. E.g. for (0,0,1) = 1/8. Should I multiply this with 9/64, so the value is 9/512 & sum up over these values? If so, every value would be less then 1/2 and the result would be 0 + 9/512 + 9/256 + 9/128 + 9/256 + 15/512 + 9/512 + 7/512 = 121/512. Or should I just sum up over probabilties like 0 +9/64 + 9/64 + 9/64 + 3/64 + 3/64 + 3/64 + 1/64 = 1 ? It wouldn't make sense for me, since every value is less then b = 1/2, b = 3/8, b = 9/16 and b = 2, I would sum up over everything and get the same result for each task. – Marceline Nov 28 '19 at 00:41
  • Let me take a specific example. F(b=3/8). Which values of Y are $\le$ 3/8? We have: (0,0,0)-->Y=0, (0,0,1) --> Y=1/8, (0,1,0) --> Y=1/4 and (0,1,1)-->Y=3/8. So, now add the probabilities of (0,0,0), (0,0,1), (0,1,0), and (0,1,1). This will be 27/64+9/64+9/64+3/64 = 48/64. So, F(b=3/8)=48/64. I hope this helps. – ad2004 Nov 28 '19 at 00:49