random sample from a geometric distribution

**rebghb** · May 15th 2010, 04:01 AM

Hello everyone!

I've got this problem on multivariable distributions I'm stuck on...

Let $X_1;X_2;X_3$ denote a random sample of size n = 3 from a distribution with the geometric pdf
$P(X=k)=(\frac{3}{4}) ( \frac{1}{4} )^{k-1}$ and given the distribution $Z = X_1 + X_2 + X_3$
Question find the pdf of $Z$ then find $P(Z=15)$ .
Now I considered using the moment generating function for 3 reasons: (a) $X_1, X_2, \text{and} X_3$ are a random sample size $\rightarrow \ X_1, X_2, \text{and } X_3$ are independent (b) they have the same pdf thus mgf, and (c) well, mgf can be used for this.
So i started out like this:
$M_Z (t) = E(e^{tZ})=E(e^{(X_1+X_2+X_3 )\cdot t})=E(e^{X_1 t} \cdot e^{X_2 t} \cdot e^{X_3 t})= ( E(e^{X_i t}) )^3$ $= (\frac{p e^t}{1-(1-p) e^t})^3$

Then I got stuck there... I presume I don't know how to "transform" variables in descrete distributions unlike continuous ones.
So any help is much appreciated!

**awkward** · May 15th 2010, 10:36 AM

Originally Posted by rebghb

Hello everyone!

I've got this problem on multivariable distributions I'm stuck on...

Let $X_1;X_2;X_3$ denote a random sample of size n = 3 from a distribution with the geometric pdf
$P(X=k)=(\frac{3}{4}) ( \frac{1}{4} )^{k-1}$ and given the distribution $Z = X_1 + X_2 + X_3$
Question find the pdf of $Z$ then find $P(Z=15)$ .
Now I considered using the moment generating function for 3 reasons: (a) $X_1, X_2, \text{and} X_3$ are a random sample size $\rightarrow \ X_1, X_2, \text{and } X_3$ are independent (b) they have the same pdf thus mgf, and (c) well, mgf can be used for this.
So i started out like this:
$M_Z (t) = E(e^{tZ})=E(e^{(X_1+X_2+X_3 )\cdot t})=E(e^{X_1 t} \cdot e^{X_2 t} \cdot e^{X_3 t})= ( E(e^{X_i t}) )^3$ $= (\frac{p e^t}{1-(1-p) e^t})^3$

Then I got stuck there... I presume I don't know how to "transform" variables in descrete distributions unlike continuous ones.
So any help is much appreciated!

Try comparing your moment generating function with the MGF of a negative binomial distribution.

**rebghb** · May 15th 2010, 01:43 PM

I tried but there's no exponential in the denominator, I looked up the binomial mgf too... Still, no success

**rebghb** · May 15th 2010, 02:04 PM

Ok, is it right to say:
$\frac{0.75^3e^{3t}}{(1-0.25e^{3t})^3}$ $=27\cdot (\frac{0.25}{1-0.25e^{3t}})^3 \ e^{3t}$ $=27 C_{k+3+3-1}^{3-1}(0.25)^3(0.75)^k$ $=27 C_{k+5}^{4}(0.25)^3(0.75)^k$

**awkward** · May 15th 2010, 04:07 PM

Originally Posted by rebghb

I tried but there's no exponential in the denominator, I looked up the binomial mgf too... Still, no success

Unfortunately, there is no agreement on how to define the geometrical and negative binomial distributions. You have to use a consistent set of definitions if you want your MGFs to match.

Try defining the geometrical distribution as
$f(x) = (1-p)^x p$ for $x=0,1,2, \dots$ .
That should get rid of the exponential in the denominator of the MGF. Then I think you will have a match with your definition of the negative binomial.

Another possible source of confusion is whether p is the probability of success or the probability of failure. Again, you need a consistent set of definitions in order for everything to work out.

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