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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!

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  • $\begingroup$ Are you familiar with conditional expected value? $\endgroup$
    – Presage
    Jun 20, 2019 at 21:38
  • $\begingroup$ @presage yes, the second part requires that? $\endgroup$
    – Fred
    Jun 20, 2019 at 21:39
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    $\begingroup$ 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 $\endgroup$
    – Presage
    Jun 20, 2019 at 21:42

2 Answers 2

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

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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.

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