# U~$(0,i)$ where $i$ is a fair dice roll.

If I throw a fair dice and the result is $$i$$ then I choose a point $$X$$~$$(0,i)$$

1st) What is the expected value and standard deviation of $$X$$?

2nd) if $$X > 3$$ then what is the probability that the result of the dice was $$6$$?

I think I got the 1st part correct but the second part I am confused on what method to use.

My attempt:

1st)

expected value of a dice roll = $$\frac{1}{6} *(1+2+3+4+5+6) = 3.5$$

so, $$i = 3.5$$ and $$X$$~Uni$$(0,3.5)$$

By definition of the Uniform distribution :

$$E(X)=\frac{3.5}{2}=1.75$$ $$D(X)= \frac{3.5}{\sqrt{12}}$$ 2nd) This is the part am not sure of:

if $$X$$~Uni$$(0,3.5)$$ and $$P(X>3)= 1-P(X \leq3) = 1- \frac{3}{3.5}= \frac{1}{7}$$

we want $$P(i=6|X>3) = \frac{P(i=6,X>3)}{P(X>3)}$$but idk how to get the numerator here, also I may be completely wrong about both parts so please point that out if it happens to be the case!

ALso, this might be right: $$P(i=6,X>3) = P(X>3|i=6)* P(i = 6) = \frac{3}{6}* \frac{1}{6}$$ This would imply $$P(i=6|X>3) = \frac{1/6 * 1/2}{1/7} = \frac{7}{12}$$

Any help is appreciated!

• Are you familiar with conditional expected value? Jun 20, 2019 at 21:38
• @presage yes, the second part requires that?
– Fred
Jun 20, 2019 at 21:39
• I think so, at least for the P(X>3) part. I'd rather said that to formalize your approach to the first part, you need to use conditional expectation as well Jun 20, 2019 at 21:42

a) Note that, for any $$\sigma$$-field $$\mathcal G$$, we have $$E[X] = E[E[X|\mathcal G]]$$

Let us consider $$\mathcal G$$ = $$\sigma(Y)$$, where Y is rv. that characterises the number we've rolled. Due to Y being discrete, we only need to find $$E[X|Y=k]$$ to know $$E[X|Y]$$. So take $$k\in\{1,2,3,4,5,6\}$$

$$E[X|Y=k]$$ = $$\frac{E[X\cdot \chi _{\{Y=k\}}]}{P(Y=k)}$$ = $$\frac{E[X_k*\chi _{\{Y=k\}}]}{\frac{1}{6}}$$ = $$E[X_k]$$ = $$\frac{k}{2}$$, where $$\chi$$ is the characteristic function, $$X_k$$ is the uniform rv on $$[0,k]$$ (third equality is due to independence), so

$$E[E[X|Y]]$$ = $$E[\frac{Y}{2}]$$ = $$\frac{7}{4}$$

To find $$D(X)$$ note that, $$D(X) = \sqrt{ E[X^2] - (E[X])^2 }$$, and that we already have $$E[X]$$, to find $$E[X^2]$$, we proceed similarly, considering $$E[X^2|Y=k]$$ we find it is equal to $$E[X_k^2]$$ = $$\frac{1}{k} \cdot \frac{k^3}{3}$$ = $$\frac{k^2}{3}$$, so again $$E[X^2] = E[E[X^2|Y]] = E[\frac{Y^2}{3}] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{18} = \frac{91}{18}$$, so plugging that: $$D(X) = \sqrt{ \frac{91}{18} - \frac{49}{16} }$$=$$\sqrt{\frac{91\cdot 8 - 49 \cdot 9}{16\cdot9}} = \frac{\sqrt{287}}{12}$$

b) As you noted: $$P(Y = 6 | X>3) = \frac{P(X>3)|Y=6)P(Y=6)}{P(X>3)}$$

$$P(X>3|Y=6) = P(X_6 >3) = \frac{1}{2}$$

$$P(Y=6) = \frac{1}{6}$$

The problematic one is $$P(X>3) = \sum_{k=1}^6 P(X>3|Y=k)P(Y=k) = \frac{1}{6}\sum_{k=4}^6 P(X_k>3) = \frac{1}{6}\sum_{k=4}^6\frac{k-3}{k}$$ =$$\frac{1}{6}(\frac{1}{4} + \frac{2}{5} + \frac{1}{2}) = \frac{1}{6}(\frac{5+4+10}{20}) = \frac{19}{120}$$

So, $$P(X>3) = \frac{\frac{1}{6}\cdot \frac{1}{2}}{\frac{19}{120}} = \frac{10}{19}$$

You're confusing the expected value of a dice roll with the expected value of the random variable. For example, if you throw the dice and the result is $$1$$, then $$X \sim U[0, 1]$$ and $$E(X) = 1/2$$. However, for a general $$i$$, then, as already stated in the problem, $$X \sim U[0, i] \Rightarrow E(X) = i/2$$.

Alternatively, you could suppose $$i = 1$$, then $$i = 2$$, then $$i = 3$$, ..., and repeat the same calculations until you exaust all $$6$$ possibilities.