How to calculate an expected value of maximum random variable function

Question

1. Let $X \sim U(0,1)$
   $Y=\max(X,0.5)$
   $Z=\max(X-0.5,0)$
   $W=\max(0.5-X,0)$

ask how to calculate $E(Y)$, $E(Z)$, $E(W)$

Answer 1

Written to elaborate the already explanatory comment of Dilip Sawarte

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Let $X$ be a random variable uniformly distributed on $(0,1)$. This means that $$g_X(x)=1 ~~\text{for}~~ x \in (0,1)$$

We are interested in the expectation of the random variable, $Y=\max\left(X,\dfrac{1}{2}\right)$.

Now, note that $$\begin{align}\mathbb E(f(X))&=\int_{-\infty}^\infty f(x)g_X(x) \rm{d}x\\&=\int_{0}^1f(x)\mathrm dx\\&=\int_0^{\frac 1 2}\dfrac{1}{2}\mathrm dx+\int_{\frac 1 2}^1x~~\rm dx\\&=\dfrac 1 4+\dfrac 1 2-\dfrac 1 8\\&=\dfrac 5 8\end{align}$$

Similarly other integrals can be evaluated.

I'll leave only the answers in case you needed to check:

For (b) $\dfrac{1}{8}$

For (c) $\dfrac{1}{8}$

As Dilip Sawarte points out, some graphs you'll find useful are:

for (a):

$\hspace{1 in}$

for (b):

$\hspace{1 in}$

for (c):

$\hspace{1 in}$

Note that the area of the shaded region is the expectation you're in need of!

Answer 2

Consider using the formula below with $T$ being the statement that $X>0.5$,

$$ \mathbb{E}(X)=P(T)\;\mathbb{E}(X \;|\; T)+P(\text{not } T)\; \mathbb{E}(X \;|\; \text{not }T).$$

For example, for the first case we have

$$\mathbb{E}(Y)=(1/2)\; (3/4)+(1/2)\; (1/2)=5/8.$$