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Let $X_1, X_2, X_3, Y_1, Y_2, Y_3, Z_1, Z_2, Z_3$ be random variables which have uniform distribution between 0 and 1. It means, the average of $X_1 = 0.5$

Let: $X=X_1 + X_2 + X_3,$ $Y=Y_1 + Y_2 + Y_3$, $Z=Z_1 + Z_2 + Z_3$

In this case, the probability of $\{X$ is bigger than $Y$ and $Z$ both$\}$ would be $\dfrac{1}{3}$.

My question is:

What is the probability of "$c+X$ is bigger than $Y$ and $Z$" when $c$ is a constant"?

For example: what is $\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$?

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2 Answers 2

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HINT

  1. Let $X = X_1+X_2+X_3$. What is the distribution of $X$? Well, a direct approach would be to find $$F_X(x) = \mathbb{P}[X_1+X_2+X_3 < x] = \iiint_{[0,1]^3} \mathbb{I}_{[a+b+c<x]} dadbdc,$$ which can be translated to a regular volume if you restrict the region of integration so the indicator is always 1.
  2. Then, $Y = Y_1+Y_2+Y_3$ and $Z = Z_1+Z_2+Z_3$ are defined analogously and have the same distribution with pdf $f(x) = F'(x)$. It's easy to see $f(x)$ only has support for $x \in [0,3]$.
  3. You want $$ \begin{split} \mathbb{P}\left[c+X >\max\{Y,Z\}\right] &= \iiint_{[0,3]^3} \mathbb{I}_{[c+x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\ &= \iiint_{[0,3]^3} \mathbb{I}_{[c+x > y]} \mathbb{I}_{[c+x > z]} f(x)f(y)f(z)dxdydz \end{split} $$ which can be similarly manipulated...
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  • $\begingroup$ Thank you a lot! $\endgroup$
    – martian03
    Jan 28, 2019 at 2:46
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Thank you @gt6989b

At first, the answer of $F_X(x)$ is

$F(x)=\dfrac{1}{6}x^3 $ when $0\leq x<1$

$F(x)=-\dfrac{1}{3}x^3+\dfrac{3}{2}x^2-\dfrac{3}{2}x+\dfrac{1}{2} $ when $1 \leq x <2$

$F(x)=1-\dfrac{1}{6}(3-x)^3$ when $2\leq x<3$

and secondly,

$f(x)=\dfrac{1}{2}x^2$ when $0\leq x<1$

$f(x)=-x^2 +3x-\dfrac{3}{2}$ when $1\leq x <2$

$f(x)=\dfrac{1}{2}(3-x)^2$ when $2\leq x<3$

And $f(y)$ and $f(z)$ follow same way.

Hence, $$ \begin{split} \mathbb{P}\left[X >\max\{Y,Z\}\right] &= \iiint_{[0,3]^3} \mathbb{I}_{[x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\ &= \dfrac{1}{3}\ \end{split} $$ And I think I can go further. Thank you again.

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