Showing $P(X_1+X_2<1)=\frac12$ where $X_1,X_2$ are i.i.d $U(0,1)$ variables

Question

I am given that $X_1$ and $X_2$ are iid $U(0,1)$ and want to show that

$$Pr[X_1+X_2<1]=0.5$$

My approach is to evaluate

$$\int_0^1\int_0^{1-x_1}1 \quad dx_2dx_1$$

but there seems to be a geometric approach to this that significantly simplifies the answer.

May I have some assistance?

Answer 1

The geometric argument (using the uniform distribution on the unit square) looks like

Answer 2

Let me call the variables X and Y. Suppose that $X$ and $Y$ are independent, continuous random variables having probability density functions $f_X$ and $f_Y$. The cumulative distribution function of $X + Y$ is obtained as follows: $$ P(X + Y < a) = F_{X+Y}(a)\\ = \int\int_{x+y \leq a} f_X(x)f_Y(y)dxdy = \\ = \int_{-\infty}^{\infty}\int_{\infty}^{a-y} f_X(x)f_Y(y)dx dy = \\ = \int_{-\infty}^{\infty}\int_{\infty}^{a-y} f_X(x)dx f_Y(y)dy = \\ = \int_{-\infty}^{\infty} F_X(a-y)f_Y(y)dy \\ $$

Since you have $X$ and $Y$ and they are iid $U(0,1)$ The cdf of $X$ looks like: $$ F_X(x) = \begin{cases} 0,&\text{if $0 \geq x$}\\ x,&\text{if $0 \leq x < 1$}\\ 1,&\text{if $1 \leq x$}\\ \end{cases} $$ and the pdf of $Y$ like: $$ f_y(y) = \begin{cases} 1,&\text{if $0 < x < 1$}\\ 0,&\text{else}\\ \end{cases} $$

Now you can substitute the the unknown parameter $a$ for your desired value of $1$ into the cumulative distribution function of $X+Y$.

$$ P(X + Y < 1) = F_{X+Y}(1)\\ = \int_{-\infty}^{\infty} F_X(1-y)f_Y(y)dy \\ = \int_{0}^{1} (1-y)1dy \\ = \left[ y-\frac{y^2}{2}\right]_0^1 = 0.5 $$

Answer 3

Note also that if $X_1$ and $X_2$ are iid $U(0,1)$, then also $1-X_1$ and $1-X_2$ are iid $U(0,1)$. So $$ \mathbb P(X_1+X_2<1)=\mathbb P((1-X_1)+(1-X_2)<1)=\mathbb P(X_1+X_2>1). $$
Note that $\mathbb P(X_1+X_2=1)=0$ because $X_1+X_2$ have continuous distribution. Red probabilities here are equal: $$ 1=\color{red}{\mathbb P(X_1+X_2<1)} + \underbrace{\mathbb P(X_1+X_2=1)}_{0}+\color{red}{\mathbb P(X_1+X_2>1)} $$ then $\mathbb P(X_1+X_2<1)=0.5$.