Expectation of a equation with uniformly distributed variable

Question

There is the following problem: One maximises profit (v-0.5)x, where x ∈ {0,1} and v~U(0,1).

Now, in order to maximise this one chooses x=0 if v<0.5, and x=1 if v≥0.5.

Now, one has to find the expected value of (v-0.5)x, i.e. E[(v-1/2)x].

The solution uses following approach: $$ \int_{0.5}^1 (v-0.5) dv$$$$=1/8$$

Using this approach seems plausible, however, since we know that v is uniformly distributed, why can't we solve the problem as following (i don't get to the same solution): For v≥ 0.5: $$E[(v-1/2)x]=E[(v-0.5)1]=E[(v-0.5)]=E[(v)]-E[(0.5)]=E[(v)]-0.5=\frac {0.5+1}{2}-0.5=1/4$$

Can anyone help?

Answer 1

The confusion lies in the fact that the expectation calculation cannot simply ignore the probability of $v$ being less than $0.5$. It looks that way when the solution simply writes $$\int_{v=1/2}^1 v - \frac{1}{2} \, dv,$$ but in fact, the profit function is fully written as $$T(v) = \begin{cases} 0, & 0 \le v \le 1/2 \\ v - 1/2, & 1/2 < v \le 1. \end{cases}$$ Therefore, the expectation of $T(v)$ is $$\operatorname{E}[T(v)] = \int_{v=0}^1 T(v) f_V(v) \, dv = \int_{v=0}^{1/2} 0 \, dv + \int_{v=1/2}^1 v - \frac{1}{2} \, dv,$$ where $f_V(v) = 1$ is the uniform density on $[0,1]$. The provided solution simply takes a shortcut and omits showing how the integral is obtained. But this is entirely misleading, because if you work with the definition of $T(v)$, it is clear that $$\operatorname{E}[T(v)] \ne \operatorname{E}[v-1/2].$$ Instead, you must first condition on the outcome of $v$: $$\operatorname{E}[T(v)] = \operatorname{E}[T(v) \mid 0 \le v \le 1/2]\Pr[0 \le v \le 1/2] + \operatorname{E}[T(v) \mid 1/2 < v \le 1]\Pr[1/2 < v \le 1].$$ Only then can you proceed: $$\operatorname{E}[T(v)] = 0 \cdot \frac{1}{2} + \operatorname{E}[v - 1/2 \mid 1/2 < v \le 1]\cdot \frac{1}{2} = \left(\frac{1/2 + 1}{2} - \frac{1}{2}\right) \cdot \frac{1}{2} = \frac{1}{8}.$$ Your calculation omits the conditionality of $v > 1/2$ and therefore the need to weight the resulting conditional expectation by the probability of observing such an outcome for $v$.

Answer 2

$E(v-\frac 1 2)x \neq E(v-\frac 1 2) 1$. This is because $x=1$ only when $v \geq \frac 1 2$. So in your calculation you should not have $E(v-\frac 1 2)$. Instead it should be $E(v-\frac 1 2) 1_{v \geq \frac 1 2 }$.