problem on gamma distribution

**Sambit** · 10-31-2010 08:45 AM

let $X_1,X_2,......,X_n$ be 'n' i.i.d random variables, each following gamma (1,3) distribution.
show that $P(X_1+X_2+.....X_n > 3(n-\sqrt{n}) ) > \frac{1}{2}$

help please

**Dason** · 10-31-2010 09:06 AM

Just a quick question. The gamma gets parameterized in different ways. Which way are you using? Is the expected value of your gamma 3 or is it 1/3?

**Sambit** · 10-31-2010 09:12 AM

i mean the distribution is:-
$f(x)=\frac{1}{2}x^2e^{-x}, x>0$

i mean the expected value is 1/3

**Dason** · 10-31-2010 09:38 AM

Ah. Then yes you should really tell us that info because apparently your parameterization is different than the two I was thinking of. To me you have alpha and beta mixed up using the first parameterization I was thinking of. But that doesn't matter too much.

The next step is realizing that since these are independent and share the same parameters you can figure out the distribution of the sum.

**Sambit** · 10-31-2010 09:45 AM

yes the sum follows a gamma (1,3n) in my notation (that is, expectation is 1/3n). right?

**Dason** · 10-31-2010 09:50 AM

Originally Posted by Sambit

yes the sum follows a gamma (1,3n) in my notation (that is, expectation is 1/3n). right?

Well you got the sum right but your expectation doesn't look right to me. I would think the expectation is 3n.

**Sambit** · 10-31-2010 09:52 AM

i think the two of us are using different notations. i mean the sum has the distribution
$f(x)=\frac{1}{(3n-1)!}x^{3n-1}e^{-x}, x>0$

now is that what you got?

**Dason** · 10-31-2010 09:57 AM

Expectation is still 3n. http://en.wikipedia.org/wiki/Gamma_distribution. Once I saw the way you parameterize it and how the distribution looks I understand what's going on. All that's happening is you're putting what I call alpha in the second parameter spot, which is where I would put what I call beta. Wikipedia uses different parameter names but it agrees with me. The expectation is the product of the parameters in both our cases. 1*(3n) = (3n)*1 = 3n.

**Sambit** · 10-31-2010 10:01 AM

yes. sorry i was wrong. it was a mistake...

what now?

i am writing gamma(alpha, p). i mistakenly wrote the expectation as alpha/p,, which is actually p/alpha. clear?
sorry for the confusion

**Dason** · 10-31-2010 10:08 AM

No need to worry. From there I'm not exactly sure where it should go but it does simplify the problem I believe. It looks like Chebyshev's inequality might help but I can't guarantee that that'll work perfectly for you. I'd have to get out some paper and try things out myself. Have you learned chebyshev's inequality?

**Sambit** · 10-31-2010 10:10 AM

yes i have.

**Dason** · 10-31-2010 10:12 AM

See if you can massage it into the framework to apply chebyshev. Note that the standard deviation of the distribution is $\sqrt{3n}$

**Sambit** · 10-31-2010 10:29 AM

not getting.
$P(\sum X > \mu+k\sigma)<\frac{1}{1+k^2}$
here $k=-\sqrt{3}$ so we have $P(\sum X > \mu+k\sigma)<\frac{1}{4}$
i don't think it helps...

**Sambit** · 10-31-2010 10:44 AM

i have done it in a separate way:-
$P(X_1+X_2+.....X_n > 3(n-\sqrt{n}) ) > \frac{1}{2}= P(\frac{(\sum X-3n)}{\sqrt{3n}} > -\sqrt{3})=1-\Phi(-\sqrt{3})$
>1/2
this can be done, isn't it?

**BGM** · 10-31-2010 10:45 AM

http://en.wikipedia.org/wiki/An_ineq...and_the_median

Let $m$ be the median of $Y$
Here using the Chebyshev's/Jensen's Inequality, we get
$|m - \mu| < \sigma \Rightarrow \mu - \sigma < m < \mu + \sigma$

Then,
$\Pr\{Y > \mu - \sigma\} > \Pr\{Y > m\} = \frac {1} {2}$

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