For my masters thesis I have to (or at least want to) understand the proofs in the following paper:
Van den Steen, E. (2004). Rational overoptimism (and other biases). American Economic Review, 1141-1151.
The proof of the first proposition especially has been giving me a lot of trouble, as I seem to lack some prerequisites here. My question requires a bit of setup, so please bear with me.
There are $J$ agents who choose an action $a_{n}$ out of a set of $N$ actions. Each agent holds subjective beliefs about each action's probability to be a success. These beliefs have means $p_{n}^{i}$, which denotes the overall likelihood of action $a_{n}$ to succeed from agent i's perspective. These means are in turn i.i.d. drawn from an atomless distribution $G$ on $[\underline{p}, \bar{p}]$. Let $\mathcal{G}$ denote the joint distribution: $\mathcal{G} = \times_{N} G$. Denote by $Y_{1}$ the action chosen by agent 1.
The proposition in the paper claims that $E_{\mathcal{G}} [p_{Y_{1}}^{1} - p_{Y_{1}}^{j}] > 0 \text{ }\forall j \neq i$ if the number of distinct actions is larger than 1.
The proof from the paper goes as follows:
The distribution of $p_{Y_{1}}^{1}$, a first order statistic, is $G^{n}(p)$. The distribution of $p_{Y_{1}}^{j}$ is $G(p)$, since $Y_{1}$ is a randomly selected action from $j's$ perspective. By integration by parts, $E_{\mathcal{G}}[p_{Y_{1}}^{1} - p_{Y_{1}}^{j}] = \int_{\underline{p}}^{\bar{p}} G(x) - (G^{N}(x)) dx$, [... which] is stricly larger than zero when $N > 1$.
The bold part is where I do not know what to do. It seems to me that what I have to do is to calculate:
$E_{\mathcal{G}}[p_{Y_{1}}^{1} - p_{Y_{1}}^{j}] = \int_{\underline{p}}^{\bar{p}} \int_{\underline{p}}^{\bar{p}} (p_{Y_{1}}^{1} - p_{Y_{1}}^{j}) \mathcal{g}(p_{Y_{1}}^{1}, p_{Y_{1}}^{j}) dp_{Y_{1}}^{1} dp_{Y_{1}}^{j}$
or
$E_{\mathcal{G}}[p_{Y_{1}}^{1} - p_{Y_{1}}^{j}] = \int_{\underline{p}}^{\bar{p}} \int_{\underline{p}}^{\bar{p}} (x - y) \mathcal{g}(x, y) dx dy$
where $g(x,y)$ is the joint pdf. But I must admit that I really do not know what to do with this from there on - and am already unsure about this step. So how would I go about solving this?`In particular, where do the distributions of the two variables come into play?
I have given this some more thought, maybe someone can tell me whether I am going in the right direction here.
So the random variables are independently distributed, therefore the joint CDF of them would have to be the product of their CDFs, right? This means that $g(x,y) = g^{N}(x) * g(y)$. My problem thus becomes:
$$\int_{\underline{p}}^{\bar{p}} \int_{\underline{p}}^{\bar{p}} (x - y) g^{N}(x) g(y) dx dy$$
$$ = \int_{\underline{p}}^{\bar{p}} g(y) \left(\int_{\underline{p}}^{\bar{p}} x g^{N}(x) dx - y \int_{\underline{p}}^{\bar{p}} g^{N}(x)\right) dx dy$$
$$ = \int_{\underline{p}}^{\bar{p}} g(y) \int_{\underline{p}}^{\bar{p}} x g^{N}(x) dx - y g(y) dy$$
And here I am stuck again. Is the above stated calculation correct so far? I tried going on like this:
$$ \int_{\underline{p}}^{\bar{p}} g(y) \left(\left[x G^{N}(x)\right]_{\underline{p}}^{\bar{p}} - \int_{\underline{p}}^{\bar{p}} G^{n}(x) dx - y g(y)\right) dy$$
but this does not seem to get me anywhere near the result in the paper.
I have found the mistakes I made, and have come up with the solution.
As stated in the paper, the joint CDF of the two random variables is $G(x)^{n} G(y)$, since X is the maximum and Y is simply a random draw.
What I want to calculate is $$E[x - y] = \int_{\underline{p}}^{\bar{p}} \int_{\underline{p}}^{\bar{p}} (t - z) dG(t)^{N} dG(z)$$
Here I made the mistake I was stuck on. The joint pdf of the two random variables is not $g(t)^{N}g(z)$, as I originally wrongly assumed, but it is $g(z) N g(t) G(t)^{N - 1}$, as follows by simple derivation of the joint cdf.
The problem thus becomes: $$ = \int_{\underline{p}}^{\bar{p}} \int_{\underline{p}}^{\bar{p}} (t - z) g(z) N g(t) G(t)^{N-1} dz dt$$ and solving this - I omit a step by step solution for now, if anyone should be interested I will add it upon request in the comments - yields the equation given in the proof in the paper.