Mathematics Asked by zestiria on December 17, 2020

Let :

- $X_1,X_2$ independent, with same law
- $ Var(X_1)= sigma^2$, $E(X_1)=0$
- $G$ is their cumulative function
- Let $X= max(X_1,X_2)$ with cumulative function $F$

We want to show that $E(X)= int_{-infty}^{infty} [1-G(t)]G(t) dt $

My attempt :

$F(t)=P(X_1<t, X_2 <t)=P(X_1<t)P(X_2 <t)= G(t)^2$ so $E(X)= int_{-infty}^{infty} t 2 g(t) G(t)$ then do an integration by parts.

Or we can use $EZ= int_{0}^{infty}P(Z>t)dt$ but it works only if $Z>0$

$((G-G^2)^{-1})’ =- (g -2gG) frac{1}{ (G-G^2) ^2 } $

Let's prove it backwards:

$$mathbb{E}[T]=int_{-infty}^{infty}G(t)[1-G(t)]dt=int_{-infty}^{infty}G(t)dt-int_{-infty}^{infty}G^2(t)dt=$$

$$tG(t)Bigg|_{-infty}^{infty}-int_{-infty}^{infty}tg(t)dt-tG^2(t)Bigg|_{-infty}^{infty}+int_{-infty}^{infty}2t g(t)G(t)dt=$$

$$underbrace{t[G(t)-G^2(t)]Bigg|_{-infty}^{infty}}_{=0}-underbrace{mathbb{E}[X_1]}_{=0}+int_{-infty}^{infty}2t g(t)G(t)dt$$

To prove that the first addend is zero you have only to calculate the limits...it's easy

the second addend is zero as per initial statement

the third addend is the result. It matches with the same result you correctly calculated by another way...the proof is finished.

Edit: further details on limit calculus:

$$lim_{t rightarrow +infty}frac{t}{frac{1}{G-G^2}} rightarrow-frac{(G-G)^2}{g-2gG}=0$$

Answered by tommik on December 17, 2020

Let $X^+ = max(X,0)$ and $X^- = max(-X,0)$. Then $X^+ = displaystyle int_0^{infty} mathbb{1}(X>t)dt = max(a ge : X ge a)$ and $X^- = displaystyle int_{-infty}^0 mathbb{1}(X le t) dt = min(a le 0: X > a)$.

Now we have: $$mathbb{E}(X) = mathbb{E}(X^+ - X^-) = mathbb{E}(X^+) - mathbb{E}(X^-) = ldots = int_0^{infty} (1-F(x))dx - int_{-infty}^0F(x)dx$$

Now for $F(x) = G^2(x)$ we have:

$$mathbb{E}(X) = mathbb{E}(max(X_1,X_2)) = int_0^{infty}(1-G(x))(1+G(x)) dx - int_{-infty}^0 G^2(x)dx$$

From this point it's easy to get your fact.

$Hint$: we have that $$mathbb{E}(X) = int_0^infty (1-G(x))G(x) dx + int_0^infty (1-G(x))dx -int_0^infty G^2(x)dx +$$ $$+int_{-infty}^0(1-G(x))G(x)dx -int_{-infty}^0(1-G(x))G(x)dx = $$ $$=int_{-infty}^{infty} (1-G(x))G(x)dx + R$$

And you need to prove: $$R = int_0^{infty}(1-G(x) )dx - int_0^{infty} G^2(x)dx-int_{-infty}^{0} (1-G(x))G(x)dx equiv 0$$

Answered by openspace on December 17, 2020

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