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Knowing that $$\int_0^\infty x^n\exp(-x)\,dx=n!,$$ can we prove the following inequality: $$\int_0^n x^n\exp(-x) \,dx< \frac{n!}2 \;\;?$$

amWhy
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Aymane Gr
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  • Why dont you use MathJax and make the reading easy? – Green.H Jun 27 '18 at 16:29
  • what do you mean by * * exp(-x) ? – David Diaz Jun 27 '18 at 16:31
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    You may want to check out the incomplete gamma function (https://en.wikipedia.org/wiki/Incomplete_gamma_function), maybe specifically the Special Values section. – John Lou Jun 27 '18 at 16:38
  • In effect you are asked to prove the right-skewed Gamma distribution has a mean which exceeds its median – Henry Jun 27 '18 at 16:49
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    Let $f(x)=x^n\exp(-x)$. Then $f$ has a global maximum at $x=n$. It suffices to prove that $f(n+h) \ge f(n-h)$ for $0 \le h \le n$. This reduces to $(n+h)^n \ge e^{2 h} (n - h)^n$ or $$ \left(1+\frac{2h}{n-h}\right)^n \ge e^{2h} $$ but I don't see how to finish it right now. – lhf Jun 27 '18 at 17:02
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    The probability distribution $$ f_n(x),dx = \frac{ x^{n-1} e^{-x} , dx }{(n-1)!} \text{ for } x\ge0 \tag 1 $$ has expected value $n$, as seen by recalling that the expected value is $\displaystyle\int_0^\infty xf(x),dx,$ and has variance $n,$ as can be seen by recalling that the variance is the expected value of the square minus the square of the expected value, and also that the expected value of the square is $\displaystyle\int_0^\infty x^2 f(x),dx.$

    Now notice the convolution$,\ldots\qquad$

     – Michael Hardy Jun 27 '18 at 17:30
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    $\ldots\qquad (f_n * f_m) = f_{n+m}.$ The probability density function of the sum of independent random variables is the convolution of their densities. Thus $f_n(x),dx$ is the distribution of the sum of $n$ independent random variables each distributed as $e^{-x},dx.$ Hence the central limit theorem is applicable. Let $X$ be a random variable distributed as $(1).$ Then the distribution of $\dfrac{X-n}{\sqrt n}$ approaches $$ \frac 1 {\sqrt{2\pi}} e^{-x^2/2} , dx $$ as $n\to\infty.$ Thus $$ \lim_{n\to\infty} \int_0^n \frac{x^{n-1} e^{-x} , dx}{(n-1)!} = \frac 1 2. $$ – Michael Hardy Jun 27 '18 at 17:30
  • Your proposed integral is $\displaystyle \int_0^{n-1} \frac{x^{n-1} e^{-x} , dx}{(n-1)!},$ so it's slightly smaller than something that approaches $1/2.$ But that does not prove it's $<1/2$ (if it did I'd have made this an answer rather than a comment. – Michael Hardy Jun 27 '18 at 17:31
  • The indefinite integral can be calculated. For large n it is approximately $n!\sum_{k=0}^n\frac{1}{k!}\approx \frac{n!}{e}\lt\frac{n!}{2}$. – herb steinberg Jun 27 '18 at 19:58
  • error in previous comment by self - ignore – herb steinberg Jun 27 '18 at 20:34
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    A direct calculation is $n!(1-e^{-n}\sum_{k=0}^n\frac{n^k}{k!})=n!e^{-n}\sum_{k=n+1}^{\infty}\frac{n^k}{k!}$. – herb steinberg Jun 27 '18 at 20:44

1 Answers1

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Let's take a step back and think a bit more elementary.

Consider the graph of $f(x) = x^n \exp(-x)$. Flip it over the line $x=n$ to get $f(2n-x)$. It is enough to show that

$$\int_{n}^{2n} f(2n-x) \, dx < \int_{n}^{2n} f(x) \, dx \,.$$

I will now show that if $x \in (n, 2n)$, then $f(2n-x) < f(x)$, which is enough to prove the above. Rearranging,

$$ \begin{align} (2n-x)^n \exp(x-2n) &< x^n \exp(-x) \\ \left( \frac{2n-x}{x} \right) &< \exp(2(n-x)). \end{align} $$

Now, let's do a little shift to $x \in (0, n)$ by doing $x \mapsto x-n$. Then we need to prove that

$$\left(\frac{n-x}{n+x}\right)^n < \exp(-2x).$$

Take the logarithm of both sides(it is well-defined) to get

$$n(\ln(n-x) - \ln(n+x)) < -2x$$

To prove the above inequality, let $a(x) = n(\ln(n-x) - \ln(n+x))$. I will now show that $a'(x) < -2$ for $x \in (0,n)$ which is enough to prove the above. Computing the derivative gives

$$a'(x) = -2 \left(\frac{n^2}{n^2-x^2}\right) < -2 \left(\frac{n^2-x^2}{n^2-x^2}\right) = -2.$$