# Expected value minimum of discrete and continuous uniform distribution

I need the expected value of a minimum, normally this is not a problem at all but in this case it is the minimum of a uniform discrete and a uniform continuous distribution.

Let $X$ be a random variable from a uniform distributon over $[1,3]$ and let $Y$ be a random variable with $$p(y=1)=p(y=2)=p(y=3)=\frac{1}{3}$$ It immediately follows that $$E[X] = E[Y] = 2$$ Now I need $E[\min(X,Y)]$.

I found a lot of topics about the expected value of the minimum. However, in all topics they used only discrete or only continuous and not both.

• With Mathematica, evaluating Mean[TransformedDistribution[Min[x,y],{Distributed[x, UniformDistribution[{1,3}]], Distributed[y, DiscreteUniformDistribution[{1,3}]]}]] gives $\frac{19}{12}$. Apr 14 '14 at 15:49

For every independent nonnegative random variables $X$ and $Y$, the random variable $Z=\min(X,Y)$ has expectation $$E[Z]=\int_0^\infty P[Z\geqslant t]\,\mathrm dt =\int_0^\infty P[X\geqslant t]\,P[Y\geqslant t]\,\mathrm dt.$$ In your case, one can decompose the integral on the RHS as the sum of the integrals on the intervals $(0,1)$, $(1,2)$, and $(2,3)$, and compute $P[X\geqslant t]$ and $P[Y\geqslant t]$ separately for $t$ in each of these intervals. This yields:
• If $t$ is in $(0,1)$, then $P[X\geqslant t]=P[Y\geqslant t]=1$
• If $t$ is in $(1,2)$, then $P[X\geqslant t]=\frac{3-t}2$ and $P[Y\geqslant t]=\frac23$
• If $t$ is in $(2,3)$, then $P[X\geqslant t]=\frac{3-t}2$ and $P[Y\geqslant t]=\frac13$
Thus, $E[Z]=$ $1$ $+$ $______$ $+$ $______$ $=$ $______$.