# 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!

• One way is to use Chebyshev's inequality. That suggests you consider what the mean and SD of a $\Gamma(n,3)$ distribution are.
– whuber
Commented Oct 21, 2020 at 15:49
• Commented Oct 22, 2020 at 10:22
• @quester Sure. Any relevant inequality will do. And there are many other lines of attack. For instance, because when $n$ is integral the distribution equals the convolution of $n$ iid exponential distributions, applying the CLT instantly gives the answer. Alternatively, one can develop simple estimates for the integrand (using, say, a saddlepoint approximation) and directly demonstrate the inequality using no statistical insight at all.
– whuber
Commented Oct 22, 2020 at 13:07

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: $$E(X^q)/a^q = (1/3)^q\Gamma(n + q)/\Gamma(n)*2^q/n^q =\\ = (\frac{2}{3})^q \frac{\Gamma(n+q)}{\Gamma(n)n^q}\stackrel{for\ q \infty}\rightarrow 0$$ $$P(X < a) \ge 1 - E(X^q)/a^q \ge 1 -(\frac{10}{12})^q$$ 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.