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Let $X$ have gamma distribution with parameters $\alpha=7$ and $\lambda=1$. Investigate the value of $F_X(10)$ using these methods:

  • Find a lower bound using Chebychev's inequality.
  • Approximate the value using the central limit theorem.
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If this is homework, it should be marked as such. What have you tried so far? Do you know what Chebychev's inequality says? Do you know what the Central Limit Theorem says, and why it is (sort of) applicable in this case? – Jonathan Christensen Nov 30 '12 at 18:03
I think I figured out finding the lower bound using Chebychev's inequality. The central limit theorem deals with zn converging with standard normal distribution. I thought I had to calculate zn, which means I need n. I don't understand what the value of n would be. – Nerl Nov 30 '12 at 18:12
I used the equation z=mean of X-mu/(standard deviation/sqrt(n)) but I'm not sure what n would be. – Nerl Nov 30 '12 at 18:14
In what way can you think of your Gamma-distributed random variable as a sum of iid random variables? – Jonathan Christensen Nov 30 '12 at 18:17
Well, if W_i is the waiting time for the ith arrival, and X is the gamma random variable, X is the sum of w_i with i=1,2,...,7 but I'm still not sure how to use this – Nerl Nov 30 '12 at 18:24

The central limit theorem says that the average of iid random variables $y_i$ having mean $\mu$ and variance $\sigma^2$ converges to a normal distribution with mean $\mu$ and variance $\sigma^2/n$.

You know that we can express the gamma-distribution random variable $X$ as $$X = \sum_{i=1}^7 W_i $$ where $W_i \sim \operatorname{Exp}(1)$ has mean 1 and variance 1. That's a sum, not an average, but we can divide both sides by $7$ and then apply the CLT: $$X/7 = \frac{1}{7}\sum_{i=1}^7 W_i \; \dot\sim \; \operatorname{Normal}(1,1/7).$$ So, multiplying by 7 again, $$X \; \dot\sim \; \operatorname{Normal}(7,7).$$ Now you can use the normal CDF to approximate $F_X(10)$.

Note that we could have decomposed the Gamma into a sum of $k$ iid $\operatorname{Gamma}(7/k,1)$ random variables for any $k \in \mathbb{N}$ (this property of a distribution is called "infinite divisibility"). We could have ended up with a sum of 16 billion $W_i$ instead of just 7, but our end result would have been exactly the same. This illustrates the point that there isn't a magic number (30, often) after which the CLT "applies."

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Apart from the (obvious) fact that these are both in $[0,1]$, there is no guarantee that the approximation of $F_X(10)$ by $F_Y(10)$ with $Y$ normal $(7,1)$ (and not $(1,7)$...) is good or not. – Did Dec 2 '12 at 13:16
Thanks for catching that the mean should be $7$, not $1$, but it should be a $\operatorname{Normal}(7,7)$, not $(7,1)$. That there's no guarantee the approximation is good is exactly the point I was making in my last paragraph, but this is the approximation the problem asked for. – Jonathan Christensen Dec 3 '12 at 15:13

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