Comparing uniform random variables.

Question

$X$ is a uniformly distributed random variable on $(0,1)$

$Y$ is a uniformly distributed random variable on $(0,2)$

$Z$ is a uniformly distributed random variable on $(0,4)$

What is the probability that $X+Y>Z$?

$P(X+Y>Z)=P(Z-X-Y<0)$ If I had the distribution of $Z-X-Y$, then this is its value at zero. How to obtain the distribution of a sum? Someone told me about using convolution, or characteristic functions.

Could it be done geometrically? We look in the box $[0,1]\times[0,2]\times[0,4]$ the volume of the points satisfying $Z-X-Y<0$ and divide this volume by the volume $8$ of the box. How can this geometric approach be justified?

Answer 1

We need the volume of $\{(X,Y,Z):\ Z<X+Y\}$ inside $[0,1]\times[0,2]\times[0,4]$. This is equal to

$$\int_{0}^{2}\int_{0}^{1}(x+y)dxdy=\int_{0}^{2}\left(\frac{1}{2}+y\right)dy=3$$

Then divide by $1\times 2\times 4=8$ to get $$P(Z<X+Y)=\frac{3}{8}$$

The reason why this works is that the definition of uniform distribution in an interval is the usual length measure divided by the total length of the interval. Moreover, the distribution of a tuple of independent random variables is equal to the product measure, the usual volume divided by the total volume.

Answer 2

\begin{align*} &P(X+Y<Z)\\ &=\int_0^4 P(X+Y<z)\cdot \left( \frac{1}{4} \right)dz \\ &=\frac{1}{4} \left[\int_0^1 P(X+Y<z)dz+\int_1^2 P(X+Y<z)dz+\int_2^3 P(X+Y<z)dz+\int_3^4 P(X+Y<z)dz \right]\\ &=\frac{1}{4} \left[\int_0^1 \left(\frac{z^2}{2}\right) dz +\int_1^2 \left(z-1+\frac{1}{2}\right)dz +\int_2^3 \left(2-\frac{(3-z)^2}{2}\right)dz+\int_3^4 \left(1\right) dz \right]\\ &=\frac{1}{4} \left[\frac{1}{6} +1 +\frac{5}{6}+1 \right]\\ &=\frac{3}{4} \end{align*} Then $P(X+Y>Z)=1-P(X+Y<Z)=\frac{1}{4}$.