Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
What do you mean by "to the greatest value"? – Calvin Lin Feb 24 '14 at 6:26

2 Answers 2

up vote 1 down vote accepted

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, $$ P(X_1+X_2\leqslant X_3)=\int_0^1\frac12x_3^2\mathrm dx_3=\frac16. $$

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.