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!
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.
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
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.
