1
$\begingroup$

I'm looking at question 4 section 3 from this problem set, bottom of page 2.

Repeated (more succinctly) here:

There is a prize $V \sim \text{Unif}[0, 1]$ measured in millions of dollars. You can choose to bid any amount $b$ (in millions of dollars). If $b < \frac{2}{3}V$, then the bid is rejected and nothing is gained or lost. If $b \geq \frac{2}{3}V$, then the bid is accepted and your net payoff is $V-b$.

What is your optimal bid $b$ (to maximize the expected payoff)?

Let's call the payoff $X$.

I have solved this in two ways, both inspired by drawing a picture of the payoff. Sparing too many details one way it occurred to me to solve it was by calculating:

$$\int_0^{\frac{3b}{2}}(V-b) \, dV$$

but having read the solution (which I found confusing) I'm struggling to translate the above "intuitive" quantity into slightly more formal conditional expectation notation.

In other words: I knew that calculating the above (simple) integral was the right thing to do but unless I can formalise why it was a valid thing to do I'm wondering if it was an accident or not, and am unsatisfied.

So my question is: why is the expected payoff, $E[X]$, equal to the above integral?

$\endgroup$

1 Answer 1

1
$\begingroup$

Note that, for a fixed $b$, $X$ is a function of $V$, and moreover it is completely determined by $V$. So I'll write $X(V)$ to clarify this fact.

By the law of the unconscious statistician, $$E[X(V)] = \int X(v) f_{V}(v) \, dv,$$ where $f_{V}$ is the density of $V$. Plugging in the fact that this density is uniform over $[0,1]$, we have $$E[X(V)] = \int X(v) f_{V}(v) \, dv$$ $$= \int_{0}^{1} X(v) \, dv$$ $$= \int_{\substack{b < \frac{2}{3} v\\ 0 \leq v \leq 1}} X(v) \, dv + \int_{\substack{b \geq \frac{2}{3} v\\ 0 \leq v \leq 1}} X(v) \, dv$$ $$ = \int_{\substack{\frac{3}{2} b < v\\ 0 \leq v \leq 1}} X(v) \, dv + \int_{\substack{\frac{3}{2} b \geq v\\ 0 \leq v \leq 1}} X(v) \, dv.$$ Assuming that $0 \leq \frac{3}{2} b \leq 1$ this becomes $$ \int_{\frac{3}{2} b}^{1} X(v) \, dv + \int_{0}^{\frac{3}{2}b} X(v) \, dv,$$ which simplifies to $$ \int_{\frac{3}{2}b}^{1} 0 \, dv + \int_{0}^{\frac{3}{2} b} (v-b) \, dv = \int_{0}^{\frac{3}{2} b} (v-b) \, dv $$ from the condition you provided governing the payoff.

Finally, we return to consider the case that $0 \leq \frac{3}{2} b \leq 1$ can be violated, in this case we have $$\int_{0}^{\min\{1, \frac{3}{2} b\}} (v-b) \, dv$$ by similar reasoning.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.