# Computation of $\mathbb{E}[\min(U+W,V+W)]$

Our exam today contained a few things I wasn't able to compute starting with this:

Let U,V,W,Y,Z be independent random variables with the following distributions

• $U\sim\text{Exp}(1/5)$
• $V\sim\text{Exp}(2/3)$
• $W\sim\mathcal U(1, 5)$ (uniform distribution on $[1;5]$)
• $Y\sim\mathcal N(-13, 1)$
• $Z\sim\mathcal N(0, 5)$

Determine the following expected value: $$\mathbb{E}[(Z+\min(U+W,V+W))\cdot(Y+Z)]$$

Hint. With a suitable approach you do not need to derive any density or distribution functions by hand. All necessary means can be computed directly via the parameters of the distributions.

As mentioned in the title I failed to deduce $\mathbb{E}[\min(U+W,V+W)]$ while being able to split evrything up based on independence and linearity of expectation. I started with a substitution $T:=U+W$ and $S:=V+W$ such that with convolution I get something like this (not formally correct, just random formulas as there were too much variables for me...)

$$\mathbb{E}[\min(U+W,V+W)]=\\=\iint\limits_{-\infty}^\infty\left(\min(t,s)\right)\cdot\underbrace{\left(\int\limits_{-\infty}^\infty 1/5\cdot\exp\left(-1/5\cdot t\right)\cdot 1/4\cdot \mathbb 1_{[1;5]}(t) \right)}_{\text{pdf of }T}\cdot\underbrace{\left(\ldots\right)}_{\text{pdf of }S}\;\mathrm dt\;\mathrm ds$$

The most ridiculous thing was $W$ - I haven't had ANY clue at all, how to handle it. Can anyone explain me a more suitable approach?

EDIT: $\mathbb 1_A(x)$ denotes the indicator function which returns 1 if $x\in A$; otherwise, 0.

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Is $\min(U+W, V+W)$ the same as $\min(U,V)+W$? What is the distribution of $\min(U,V)$ since they both are independent exponential random variables? – Dilip Sarwate Aug 7 '12 at 18:39
Integrals are linear, so is minimum, so $\int(min(U+W,V+W)=\int(min(U,V)+W)=\int(min(U,V)+\int W$. – tomasz Aug 7 '12 at 18:39
@tomasz : Could you write $\min$ instead of $min$? That doesn't only de-italize it, but also provides for proper spacing in things like "$5\min X$", and affects positioning of subscripts when things like $\displaystyle\min_{x\in A}$ appear in a "displayed" setting. – Michael Hardy Aug 7 '12 at 19:24
It's true that $\min(U+W,V+W) = \min(U,V)+W$, but that doesn't make minimum linear. – Robert Israel Aug 7 '12 at 19:33
@RobertIsrael: touché. It is until we multiply by negative numbers. ;) Still, it is additive, which is all we need. – tomasz Aug 7 '12 at 19:48

Since $\min(U+W,V+W)=\min(U,V)+W$, one is left with computing $\mathrm E(M)$ when $M=\min(U,V)$ and the random variables $U$ and $V$ are independent and exponential with respective parameters $a$ and $b$.
In this specific case, the distribution of $M$ is quite simple: consider that the distribution of $U$ is characterized by the fact that $\mathrm P(U\geqslant x)=\mathrm e^{-ax}$ for every $x\geqslant0$, that the distribution of $V$ is characterized by the fact that $\mathrm P(V\geqslant x)=\mathrm e^{-bx}$ for every $x\geqslant0$, and that, for every $x\geqslant0$, $$[M\geqslant x]=[U\geqslant x]\cap[V\geqslant x].$$ By independence of $U$ and $V$, $\mathrm P(M\geqslant x)=\mathrm e^{-ax}\cdot\mathrm e^{-bx}=\mathrm e^{-(a+b)x}$ for every $x\geqslant0$ hence the distribution of $M$ is exponential with parameter $a+b$. In particular $\mathrm E(M)=1/(a+b)$.