1. mgf

$\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...} \\ 0 & \mbox{ elsewhere } \end{array}\right. \\ \mbox{ where } 0<p<1 \\$

how do I show that

i)$\displaystyle M_x(t)=\frac{pe^t}{1-(1-p)e^t}$
ii)use the mgf to find the mean and variance of X and

iii) find P(X + Y = 2)?

I'm lost with mgf's, I've searched the web but just definitions are given so I'm hoping to find examples

2. Hello,

The MGF of a random variable X is :
$\displaystyle M_X(t)=\mathbb{E}(e^{tX})$

For discrete distributions, we have $\displaystyle \mathbb{E}(g(X))=\sum_{k=1}^\infty g(k) \mathbb{P}(X=k)$

So here, we have :

\displaystyle \begin{aligned}M_X(t) &=\sum_{k=1}^\infty e^{tk} \mathbb{P}(X=k) \\ &=\sum_{k=1}^\infty e^{tk}p(1-p)^{k-1} \\ &=e^t \sum_{k=1}^\infty e^{t(k-1)} p(1-p)^{k-1} \end{aligned}

Now change the indice so that it starts at k=0 :

\displaystyle \begin{aligned} M_X(t) &=e^t \sum_{k=0}^\infty e^{tk} p(1-p)^k \\ &=pe^t \sum_{k=0}^\infty (e^t(1-p))^k \end{aligned}

And this is just a geometric series

3. Thanks but how would I attempt parts (ii) and (iii)

4. Originally Posted by bigdoggy
$\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...} \\ 0 & \mbox{ elsewhere } \end{array}\right. \\ \mbox{ where } 0<p<1 \\$

how do I show that

i)$\displaystyle M_x(t)=\frac{pe^t}{1-(1-p)e^t}$

ii)use the mgf to find the mean and variance of X and

iii) find P(X + Y = 2)?
[snip]
Originally Posted by bigdoggy
Thanks but how would I attempt parts (ii) [snip]
It's called moment generating function for a reason ..... Your notes must say how to use it to calculate E(X) and E(X^2) ....?

Originally Posted by bigdoggy
[snip]
iii) find P(X + Y = 2)?

I'm lost with mgf's, I've searched the web but just definitions are given so I'm hoping to find examples
What's Y? And are X and Y independent?

5. As for the steps for the mgf :

$\displaystyle e^{tk}=(e^t)^k$ (rule of exponents)

And for the final step, remember that $\displaystyle \sum_{n=0}^\infty x^n=\frac{1}{1-x}$ if $\displaystyle |x|<1$

6. Thanks for you're help moo and mr f...sorry for double posting Iam quite new to posting the questions and I thought a question about mgf and then the solving of the gp validated another thread...lesson learnt!

Mr f...sorry, X & Y are random and independent...

7. Originally Posted by bigdoggy
Thanks for you're help moo and mr f...sorry for double posting Iam quite new to posting the questions and I thought a question about mgf and then the solving of the gp validated another thread...lesson learnt!

Mr f...sorry, X & Y are random and independent...
I also asked what's Y? Re-phrasing: what distribution does Y follow? Things go much easier if you post the whole question.

8. X&Y are independent random variables with common distribution given by:

$\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...} \\ 0 & \mbox{ elsewhere } \end{array}\right. \\ \mbox{ where } 0<1<p \\$

That's the info given...

9. Originally Posted by bigdoggy
X&Y are independent random variables with common distribution given by:

$\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...} \\ 0 & \mbox{ elsewhere } \end{array}\right. \\ \mbox{ where } 0<1<p \\$

That's the info given...
The choices are
X = 0, Y = 2.
X = 1, Y = 1.
X = 2, Y = 0.

It should be simple to calculate the probability of these events. Remember to multiply for 'and' and add for 'or' ....

10. Originally Posted by mr fantastic
The choices are
X = 0, Y = 2.
X = 1, Y = 1.
X = 2, Y = 0.

It should be simple to calculate the probability of these events. Remember to multiply for 'and' and add for 'or' ....
P(X=1)*P(Y=1) = 2p, since k=1,2,... can't have P(X=0) nor P(Y=0)?