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A game is to choose a random real number $x$ between 0 and 10. The earnings are given by $|5-X|$ being X the number chosen. (a) - Find the earning distribution and (b) - If you play twice with $X_1$ $X_2$ and $X = max\{X_1,X_1\}$, what it's the earning distribution?.

Here's what i tried:

(a) - $f_X(x) = \left\{ \begin{array}{lr} 2-x & : x \leq \theta \\ x-8 & : x > \theta \\ 0&: otherwise \end{array} \right.$ , then

$F_X(\alpha <X < \beta) = \displaystyle\int_{\alpha}^{\beta}f_X(x)dx= \left\{\begin{array}{lr} 2x-\frac{1}{2}x^2|_a^{\beta} & : 0 \leqq \alpha < \beta \leq 2 \\ 2x-\frac{1}{2}x^2|_a^2 & : 0 \leq \alpha \leq 2 \leq \beta \\ \frac{1}{2}x^2-8x|_a^{\beta}&: 8 \leq \alpha < \beta \leq 10 \ \\ \frac{1}{2}x^2-8x|_8^{\beta}&: \alpha \leq 8 \leq \beta \leq 10 \\ 0 &: 2 \leq \alpha < \beta \leq 8 \end{array}\right.$.

It's ok?. Also, im not sure how to do (b).

Cure
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    Is $X_1$ discrete uniform or continuous uniform? Oddly enough, continuous us marginally simpler for the first part. In the first part, what you wrote seems taken from a solution to an entirely different problem. – André Nicolas Jul 09 '13 at 04:35
  • I believe all distributions are continuous here since the game was to choose a random real number between 0 and 10 -i forgot that part, i'll fix it right now-. Should i use the uniform distribution here?, but then how could i define a distribution for the earnings? – Cure Jul 09 '13 at 04:46
  • I will write a solution. About $10$ minutes. – André Nicolas Jul 09 '13 at 04:48

1 Answers1

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We interpret "at random" to mean with uniform distribution. In principle, however, "at random" could here mean any distribution with support $[0,10]$.

We change the notation slightly. Let $U_1$, $U_2$ be independent continuous uniform on $[0,10]$. Let $Y_i=|5-U_i|$. It is almost obvious that the $Y_i$ are uniform on $[0,5]$.

Parenthetically, note that in the discrete case, things are marginally more complicated, since then $\Pr(Y_i=0)$ is half of $\Pr(Y_i=k)$ where $k\ne 0$.

Now let $X=\max(Y_1,Y_2)$. We find the cdf $F_X(x)$ of $X$.

Of course the cdf is $0$ before $0$ and $1$ after $5$. Now we deal with $0\lt x\lt 5$.

The larger of $Y_1$ and $Y_2$ is $\le x$ iff they are both $\le x$. The probability of this is $\frac{x}{5}\cdot\frac{x}{5}=\frac{x^2}{25}$. If you want the density function, differentiate.

Added: For completeness we find the distribution of the $Y_i$. For $0\le y\le 5$, we have $Y_i\le y$ if and only if $|X-5| \le y$ if and only if $5-y\le X\le 5+y$. The length of this interval is $2y$, so the probability is $\frac{2y}{10}=\frac{y}{5}$. Thus the cdf of the $Y_i$ is $\frac{y}{5}$ on the interval $[0,5]$, and therefore the density is $\frac{1}{5}$. Thus $Y_i$ is uniform on $[0,5]$.

André Nicolas
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  • Thanks!!. Now i have a couple question about the solution: (1) What would happen if i have to pay $k$ to play the game?, the earning distribution would be $|5-U_i| - k$ ?, then the cdf is $0$ before $0$ and $1$ after $5-k$? – Cure Jul 09 '13 at 05:38
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    The cdf would of what I called the $Y_i$ would be $0$ before $-k$ and $1$ after $5-k$. We would get a uniform distribution on $[-k,5-k]$. – André Nicolas Jul 09 '13 at 05:50