0

I am trying to understand how the following function is integrated by parts (F(x) is the cdf of f(x)):

$$ \frac {\int_t^{+\infty}[1-F(x)]dx} {\int_{-\infty}^t F(x) dx} $$

After integrating by parts this becomes(I am familiar with the integrating by parts procedure but don't quite see it is applied to arrive at the equation below):

$$ \frac {\int_t^{+\infty}(x-t) f(x)dx} {\int_{-\infty}^t (t-x) f(x) dx} $$

And some more transformation (also not sure what is happening here):

$$ \frac {E[(x-t)^+]} {E[(t-x)^+]} $$

And one more (also unclear to me):

$$ \frac {E(x)-t} {E[(t-x)^+]}+1 $$

Giorgos Giapitzakis
  • 6,416
Niccola Tartaglia
  • 101
  • $1-F(x) = (1-F(x)) \cdot 1 = (1-F(x)) \cdot (x)'$ and when you integrate by parts you get $(1-F(x))' x=-xf(x)$ and boundary terms... – Giulio R Oct 07 '21 at 19:55

1 Answers1

1

Here is a direct argument. Let $X$ be a random variable with cdf $F$ s.t. $\mathsf{E}|X|<\infty$. Then, using the tail sum formula for expectations (you can find the IBP argument there), $$ \int_{t}^{\infty}(1-F(x))\,dx=\int_0^{\infty}(1-F(x+t))\,dx=\mathsf{E}[X-t]^{+}, $$ where $[X-t]^{+}=\max\{X-t,0\}$, and $$ \int_{-\infty}^t F(x)\,dx=\int_{-\infty}^0 F(x+t)\,dx=\int_0^{\infty} F(t-x)\,dx=\mathsf{E}[t-X]^{+}. $$ Thus, \begin{align} \frac{\int_{t}^{\infty}(1-F(x))\,dx}{\int_{-\infty}^t F(x)\,dx}&=\frac{\mathsf{E}[X-t]^{+}}{\mathsf{E}[t-X]^{+}} \\ &=\frac{\mathsf{E}[X-t]+\mathsf{E}[t-X]^{+}}{\mathsf{E}[t-X]^{+}} \\ &=\frac{\mathsf{E}X-t}{\mathsf{E}[t-X]^{+}}+1 \end{align} because $a=\max\{a,0\}+\min\{a,0\}=\max\{a,0\}-\max\{-a,0\}$.

  • Thank you, going through your steps now. – Niccola Tartaglia Oct 09 '21 at 16:58
  • In the first row, I understand why your substitution of x+t for x (because you changed the lower boundary of the integral). And I assume your motivation was, so that you could use the tail sum formula you mention. But how did you get to the next step. I couldn't figure this out with the forum link you shared. I am not really familiar with the tail sum formula for expectations. – Niccola Tartaglia Oct 09 '21 at 18:44
  • Nevermind, just googled it. First row is clear now. – Niccola Tartaglia Oct 09 '21 at 18:46
  • What exactly does the plus sign imply though, I figure it means anything above 0 to infinity, but I don't see that mentioned in the tail sum formula. – Niccola Tartaglia Oct 09 '21 at 18:48
  • @NiccolaTartaglia The plus sign means the positive part of a random variable. –  Oct 09 '21 at 20:05
  • Thank you for clarifying. But I am somewhat still struggling with that last step in the first line. Is there a specific version of that tail sum formula we used here? – Niccola Tartaglia Oct 10 '21 at 21:36
  • @NiccolaTartaglia It's $\int_0^{\infty}\mathsf{P}(X>x+t),dx=\int_0^{\infty}\mathsf{P}(X-t>x),dx=\mathsf{E}\max{X-t,0}$. –  Oct 10 '21 at 23:44
  • Ah, that helps. Thank you so much, appreciate the additional steps here!!!! – Niccola Tartaglia Oct 11 '21 at 15:39
  • If I may ask one more follow up. In the second to last step (you indicated that this is because of the equality you reference below), would you be able to add one more step to explain how you arrived at this: E[X-t]^+= E[X-t] +E[t-X]^+ – Niccola Tartaglia Oct 11 '21 at 16:17
  • @NiccolaTartaglia Sure. \begin{align} \mathsf{E}[X-t]&=\mathsf{E}\max{X-t,0}+\mathsf{E}\min{X-t,0} \ &=\mathsf{E}\max{X-t,0}-\mathsf{E}\max{t-X,0}. \end{align} Now, rearrange the equation. –  Oct 11 '21 at 16:25
  • I see, thank you so much, I highly appreciate the help!!! – Niccola Tartaglia Oct 16 '21 at 21:20