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I have just recently learned calculating Expected Value via Darth Vader Rule which states that $$E(X) = \int_{0}^{\infty}1-F(X) dx $$

I decided to try this for a uniform random variable $X$ defined over $[3,8]$ which, intuitively, $E(X)$ should be 5.5. However, when I tried this via Darth Vader Rule, I got a different result:

Assuming $X$ is uniform over $[3,8]$, its PDF is $f(x) = \dfrac{1}{5}$ and its CDF is $F(X)=\dfrac{x-3}{5}$

$$E(X)=\int_{3}^{8}1-\left(\dfrac{x-3}{5}\right) dx = 2.5$$

Which is a bit unexpected. I have tried this for different uniform distribution $U(a,b)$ and it seems that instead of $E(X)=\dfrac{a+b}{2}$, the Darth Vader always produces $E(X)= \dfrac{a+b}{2}-a$.

What did I do wrong here? Any comment/insight would be much appreciated.

Saito
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    Who calls the tail sum formula Darth Vader rule??? Anyway, you did not integrate from 0 to 3. (F(x) = 0 for when x \in [0, 3]); that would correct your expectation value. – E-A Nov 12 '21 at 02:49
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    Found a Math.SE post about the naming in case anyone else is curious: https://math.stackexchange.com/questions/919737/darth-vader-rule-what-is-the-reason-for-its-name-and-a-formal-proof ("So, I think the correct answer is: there is no sensible reason why the rule is called by this name --- some maths guys just thought it would be a cool name because they are Star Wars nerds." -answer by Ben) – E-A Nov 12 '21 at 02:50
  • Well, my professor called it a Darth Vader rule saying the guy called it that on a whim too. Anyway, isn't it the case that the integral must be integrated over the value of the random variable though (in this case, $X$) i.e. from 3 to 8?

    Also I did try integrating from 0 to 3 as you suggested. Still turns into 3.9.

     – Saito Nov 12 '21 at 02:53
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    Your error is that here $\int\limits_{0}^{\infty}(1-F(X)), dx = \int\limits_{0}^{3}1, dx +\int\limits_{3}^{8}\left(1-\left(\dfrac{x-3}{5}\right)\right) ,dx+\int\limits_{0}^{\infty}0,dx$ and you missed the first term – Henry Nov 12 '21 at 02:57
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    Oh! That makes perfect sense now and I finally understand what @E-A meant. Thank you all! – Saito Nov 12 '21 at 02:58
  • @Saito You may also find the following Math.SE pages useful: https://math.stackexchange.com/questions/172841/explain-why-ex-int-0-infty-1-f-x-t-dt-for-every-nonnegative-rando, and https://math.stackexchange.com/questions/64186/intuition-behind-using-complementary-cdf-to-compute-expectation-for-nonnegative – E-A Nov 12 '21 at 02:58
  • I for one love that you call this the Darth Vader rule – OskarSzarowicz Nov 12 '21 at 09:43

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