Gamma Distribution satisfying property How can we prove that gamma random variable $X_{n}$ with parameters $(n,3)$ can satisfy the following relation for some $n$?
$$P(X_{n} < n/2) > 0.999$$
I used the definition of density function for $X_{n}$, but it was hard to integrate and obtain exact value for probability. Would welcome to know your approaches!
 A: I'm pretty sure the expected answer to this problem involves an argument such
as the one suggested by @whuber.
However, because the values are specific, a grid search in R can be used to
find an exact solution, as shown below. I assume $3$ is the gamma rate parameter, as in R. [There is no general
agreement whether the second parameter of a gamma random variable is a 'rate' or 'scale' parameter, so you should always say which applies.]
n = 1:100               # hoping answer is < 101
p = pgamma(n/2, n, 3)   # P(X_n < n/2) for X_n ~ Gamma(shape=n/2, rate=3)
min(n[p > .999])        # smallest n that meets condition
[1] 50
pgamma(50/2, 50, 3)     # verification
[1] 0.9990961           # OK
pgamma(49/2, 49, 3) 
[1] 0.9989981           # not OK

Graph:
curve(dgamma(x, 50, 3), 0, 30, lwd=2, ylab="PDF", main="Gamma(shape=50, rate=3)")
 abline(h=0, col="green2")
 abline(v=0, col="green2")
 abline(v=25, col="orange", lwd=2, lty="dotted")


Note: If the scale is $3$ and rate is $1/3,$ then there is no positive integer $n$ that works:
pgamma(1/2, 1, 1/3) 
[1] 0.1535183         # too small
pgamma(2/2, 2, 1/3) 
[1] 0.04462492        # even smaller
pgamma(10/2, 10, 1/3) 
[1] 1.011967e-05      # etc.

A: from markov inequality:
$$P(X \ge a) \le E(X^q)/a^q$$
$$P(X \ge a) = 1 - P(X \lt a) \le E(X^q)/a^q => P(X < a) \ge 1 - E(X^q)/a^q$$
from
https://en.wikipedia.org/wiki/Generalized_gamma_distribution
https://en.wikipedia.org/wiki/Gamma_distribution#General
$$E(X^q) = a \Gamma((d+1)/p)/\Gamma(d/p) \stackrel{d=n/q, a=(1/3)^q, p=1/q}=(1/3)^q\Gamma(n + q)/\Gamma(n)$$
now:
$$E(X^q)/a^q = (1/3)^q\Gamma(n + q)/\Gamma(n)*2^q/n^q =\\
= (\frac{2}{3})^q \frac{\Gamma(n+q)}{\Gamma(n)n^q}\stackrel{for\ q<n/4\ and\ big\ n}<(\frac{2}{3})^q (\frac{5}{4}n)^q/n^q = (\frac{10}{12})^q \stackrel{q -> \infty}\rightarrow 0$$
$$P(X < a) \ge 1 - E(X^q)/a^q \ge 1 -(\frac{10}{12})^q$$ and two easy steps...
There are some minor mistakes (in step with dropping Gammas) but whole idea should be correct
A: The best Chebyshev inequality (see the reference below) for gamma random variables, where $k \gt E[X]$, is given by $$P \left[ X > k \right] \leq \frac{k f(k)}{k \lambda - \alpha} \ ,$$ with $f(x)$ being the density function $$f(x)=\frac{x^{\alpha-1}\lambda^\alpha e^{-\lambda x}}{\Gamma \left( \alpha \right)}$$
If we reverse your inequality, we are looking to satisfy $$P \left[ X_n \gt n/2 \right] \le 0.001 $$
Plugging in your values of $\alpha=n$, $\lambda=3$, and $k=n/2$, we want $$\frac{k f(k)}{k \lambda - \alpha}= \frac{\left(n/2 \right) \left(n/2 \right)^{n-1} 3^n e^{-3n/2}}{\left( 3n/2-n \right)   \Gamma \left( n \right)}\leq 0.001 $$
This simplifies to finding the smallest $n$ satisfying $$\frac{ \left(n/2 \right)^{n-1} 3^n e^{-3n/2}}{   \Gamma \left( n \right)}\leq 0.001 $$
This has to be solved numerically, and the result is $n=50,$ matching BruceET's solution.
Reference: Best constants in Chebyshev Inequalities with Various Applications, Anirban Dasgupta, Technical Report # 98-20, Department of Statistics, Purdue University, West Lafayette, IN, November 1998, Revised November 2000.
