Uniform Distribution in [0,1] where P[x1+x2<=x3]

Question

Consider the following question : X1, X2, X3 are 3 independent random variables having uniform distribution between [0,1] then P[x1+x2<=x3] to the greatest value is ?

Now this is not a homework. I am not thorough with the concepts of uniform distribution. So please guide me how to solve this question or provide some links where I can learn about the concepts required to solve suck questions. Thanks in advance.

Answer 1

Re "concepts": every $X_i$ has density $f=\mathbf 1_{(0,1)}$ hence, by independence, $$ P(X_1+X_2\leqslant X_3)=\iint\!\!\!\int f(x_1)f(x_2)f(x_3)\mathbf 1_{x_1+x_2\leqslant x_3}\mathrm dx_1\mathrm dx_2\mathrm dx_3. $$ Re computations: a shortcut is to note that, for every $x$ in $(0,1)$, $$ P(X_1+X_2\leqslant x)=\iint f(x_1)f(x_2)\mathbf 1_{x_1+x_2\leqslant x}\mathrm dx_1\mathrm dx_2, $$ is the area of the triangle with vertices $(0,0)$, $(x,0)$ and $(0,x)$ in the $(x_1,x_2)$-plane, that is, $\frac12x^2$. Thus, $$ P(X_1+X_2\leqslant X_3)=\int_0^1\frac12x_3^2\mathrm dx_3=\frac16. $$

Answer 2

We are given that $X_1, X_2, X_3 \sim U[0,1]$

Hint: Show $X_1 + X_2 \sim G$, where the probability distribute function is $g(x) = \begin{cases} x & 0\leq x\leq 1 \\ 2-x & 1 < x \leq 2 \\ 0 & \text{otherwise} \end{cases}$

Hint: Evaluate the cumulative distribution function $G(x) = \int^x g(y) \, dy$.

Hint: Hence, $P(X_1 + X_2 \leq X_3) = \int_0^1 G(y) \times 1 \, dy$.