Gamma distribution and probability less then expected value?

Question

Let $X\sim \operatorname{Gamma}(\alpha = 7, \beta)$, then $P(X > E(X))$ is:

A) 0.35

B) 0.45

C) 0.55

D) 0.65

The answer is 0.45.

This is what I have so far:

$E(X)=\alpha\beta$

so I want $P(X > \alpha \beta)$, $\alpha=7$ so I can use the table for this if I divide by beta, but I don't have beta's original value. From the table I'm deducting that $X=7$..

My main question is, when I divide by $\beta$, am I also dividing the mean and variance by Beta? if that's the actual case I'm guessing $$P(X > E(X))= P(X > 7)$$ which then would make sense since $\alpha$ and $x$ are both 7. So is this the general rule when dividing by beta to make it equal to 1 to use the tables?

Thank you!!! sorry for the rather lengthy question

Answer 1

You are headed in the right direction. Let's use R statistical software to try this for forty different values of $\beta:$

beta = seq(.5, 20, by=.5);  mu = 7*beta
1 - pgamma(mu, shape=7, scale=beta)
 [1] 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111
 [9] 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111
[17] 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111
[25] 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111
[33] 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111 0.4497111

From @BGM's Comment, what is the answer when $\beta = 1?$ (Second of 40 results.) Get the idea?

Answer 2

The value of $\beta$ doesn't matter at all since $\beta$ is a scale parameter, so altering $\beta$ merely multiplies both sides of the inequality $X> \operatorname{E}(X)$ by the same number. Thus for simplicity we can assume $\beta=1.$

The probability distribution is then $$ \frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x} \, dx \text{ for } x\ge 0. $$ The expected value is $\alpha.$ So \begin{align} \Pr(X>\operatorname{E}(X)) = {} & \Pr(X>\alpha) = \int_\alpha^\infty \frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x}\,dx = \frac 1 {720} \int_7^\infty x^6 \Big(e^{-x} \, dx\Big) \\[10pt] = {} & \frac 1 {720} \int u\,dv = \frac 1 {720} \left( uv - \int v\,du \right) \\[10pt] & \text{where } u = x^6 \text{ and } dv = e^{-x} \, dx. \end{align} When you do this integration by parts, you will get an expression involving $$ \int_7^\infty 6x^5 e^{-x} \, dx $$ and you will have to integrate by parts again, and iterate that until the power of $x$ is $0$.

Another way to look at it is this: Let $X$ be the waiting time until the $7$th arrival in a Poisson process. Then $X$ has this distribution. But the following events are the same: $$ \Big[ X > 7 \Big] \quad \text{and} \quad \Big[ \text{fewer than 7 arrivals before time 7} \Big] $$ The number of arrivals before time $7$ has a Poisson distribution with expected value $7$, so the probability is $$ \Pr(\text{fewer than 7 arrivals}) = \sum_{n=0}^6 \frac{7^n e^{-7}}{n!}. $$