# Do distributions weights always sum up to 1

I have been given the following problem: 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 $$\in [0,1]$$. What are the probability weights of $$Y$$.

I have been given this problem before, but I was just asked to compute the probability for $$Z$$, which is fairly straightforward looking at 4 different outcomes (0 hits = $$\frac{1}{8}$$, 1 hit = $$\frac{3}{8}$$, 2 hits = $$\frac{3}{8}$$, 3 hits = $$\frac{3}{8}$$).

1) Why am I looking at a continuous $$[0,1]$$ variable?

2) I computed the weights by just plugging in all possible combinations, but the weights don't sum up to 1. They sum up to 3.5. Why is that? And can I now just take weights that describe the same outcome (e.g. (hit, no hit, no hit), (no hit, hit, no hit), (no hit, not hit, hit)) sum them, divide by 3.5 and that's it? I have a feeling this is not how to do it.

Thanks!

/edit: There are a couple of follow up questions on my exercise sheet. I don't care too much about the exact answer (right away at least. eventually, of course I care about it :D). But really, what I want is get an understanding of the random variable I'm looking at. I'm missing something!

You may be over thinking this: there are eight possible values for the triplet $$Z$$ and each corresponds to a different $$Y$$. These are:

   Z         Y       probability
(0,0,0)      0          27/64
(0,0,1)     1/8          9/64
(0,1,0)     1/4          9/64
(0,1,1)     1/4          3/64
(1,0,0)     1/2          9/64
(1,0,1)     5/8          3/64
(1,1,0)     3/4          3/64
(1,1,1)     7/8          1/64

• Maybe.. how did you get to those numbers? For (0,0,0) I got $\frac{1}{2} \cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{8} \cdot \frac{3}{4} = \frac{21}{32}$. Yours looks like what I expected. – Marc Nov 22 '19 at 14:42
• The probability of three $0$s is $\frac34 \times \frac34 \times \frac34$ and then $Y = \frac12 \times 0 + \frac14 \times 0 + \frac18 \times 0 =0$ – Henry Nov 22 '19 at 14:49
• Okay I see now. I am overthinking this. But I got it, I can now make sense of the follow up questions. Thanks! – Marc Nov 22 '19 at 14:57
1. It's not continuous; the support is contained in $$[0,1]$$ but actually the support is the binary-represented numbers $$0.000,0.001,0.010,0.011,0.100,0.101,0.110,0.111$$. Each value of $$Z$$ corresponds to exactly one of these values of $$Y$$.
2. You have that the probability of a given value of $$Y$$ is $$1/4$$ to the power of the number of $$1$$s times $$3/4$$ to the power of the number of $$0$$s in the binary representation. You can show that the sum of all 8 of these is $$1$$ either by a direct calculation or by using the binomial theorem, since there will be $${3 \choose k}$$ numbers with $$k$$ 1s in them, resulting in $$\sum_{k=0}^3 {3 \choose k} (1/4)^k (3/4)^{3-k}=1$$.
• Yes! Thats it! I think this is what this exercise is about. I just found that out as well. Thanks Ian! Sorry I wish I could accept both answers, but Henrys approach helped more understanding the actual issue. – Marc Nov 22 '19 at 16:44