# Discrete Distributions

• Mar 11th 2009, 11:19 PM
MathRules!
Discrete Distributions
(i)Given the name of the distribution of X; (ii)find the values of ” and σ^2.

(a)M(t) = (0.3 + 0.7e^t)^5
(b)M(t) = 0.45 + 0.55e^t
(c)M(t) = 0.3e^t + 0.4e^2t + 0.2e^3t + 0.1e^4t

For (a) I said that it was a binomial distribution for part (i). Then for part (ii),
M'(t) = 5(0.3 + 0.7e^t)^4(0.7e^t)
M'(0) = 5(0.3 + 0.7e^0)^4(0.7e^0) = 5 (0.7) = 3.5

therefore ” = 3.5

M''(t) = 20(0.3 + 0.7e^t)^3(0.7e^t)^2 + 5(0.3 + 0.7e^t)^4(0.7e^t)
M''(0) = 20(0.3 + 0.7e^0)^3(0.7e^0)^2 + 5(0.3 + 0.7e^0)^4(0.7e^0) = 13.3

therefore σ^2 = M''(0) - [M'(0)]^2 = 13.3 - 3.5^2 = 1.05

For (b) I said that it was a Bernoulli distribution for part (i). Then for part (ii),
” = np = 1(0.55) = 0.55
σ^2 = np (1-p) = 0.55 ( 1- 0.55) = 0.2475

For (c) I said that it was not a usual distribution
” = M'(0) = 2.1
σ^2 = M''(0) - [M'(0)]^2 = 3.2 - 2.1 = 1.1 <---- I am very not sure here

Is my logic correct?
• Mar 11th 2009, 11:55 PM
matheagle
Quote:

Originally Posted by MathRules!
(i)Given the name of the distribution of X; (ii)find the values of ” and σ^2.

(a)M(t) = (0.3 + 0.7e^t)^5
(b)M(t) = 0.45 + 0.55e^t
(c)M(t) = 0.3e^t + 0.4e^2t + 0.2e^3t + 0.1e^4t

For (a) I said that it was a binomial distribution for part (i). Then for part (ii),
M'(t) = 5(0.3 + 0.7e^t)^4(0.7e^t)
M'(0) = 5(0.3 + 0.7e^0)^4(0.7e^0) = 5 (0.7) = 3.5

therefore ” = 3.5

M''(t) = 20(0.3 + 0.7e^t)^3(0.7e^t)^2 + 5(0.3 + 0.7e^t)^4(0.7e^t)
M''(0) = 20(0.3 + 0.7e^0)^3(0.7e^0)^2 + 5(0.3 + 0.7e^0)^4(0.7e^0) = 13.3

therefore σ^2 = M''(0) - [M'(0)]^2 = 13.3 - 3.5^2 = 1.05

For (b) I said that it was a Bernoulli distribution for part (i). Then for part (ii),
” = np = 1(0.55) = 0.55
σ^2 = np (1-p) = 0.55 ( 1- 0.55) = 0.2475

For (c) I said that it was not a usual distribution
” = M'(0) = 2.1
σ^2 = M''(0) - [M'(0)]^2 = 3.2 - 2.1 = 1.1 <---- I am very not sure here

Is my logic correct?

The MGFs are unique. So, once you noticed that
$\displaystyle M(t) = (0.3 + 0.7e^t)^5$ is a binomial with n=5 and p=.7.
Then the mean is just np=(5)(.7)=3.5 and the variance is npq....
There's no need to differentiate.

For the third MGF, you need to go back to the definition of our MGF.
I always mention these.
$\displaystyle M_X(t)=\sum_xP(X=x)e^{xt}=.3e^t + .4e^{2t} + .2e^{3t} +.1e^{4t}$
Now just match up terms. The coefficients are the probabilities and that's why they sum to one, and $\displaystyle M(0)=1$.
And the constants in front of the t in the exponent are just the realizations of x.

Thus $\displaystyle P(X=1)=.3,P(X=2)=.4, P(X=3)=.2,P(X=4)=.1$.
• Mar 12th 2009, 12:02 AM
Moo
Hello,
Quote:

Originally Posted by MathRules!
(i)Given the name of the distribution of X; (ii)find the values of ” and σ^2.

(a)M(t) = (0.3 + 0.7e^t)^5
(b)M(t) = 0.45 + 0.55e^t
(c)M(t) = 0.3e^t + 0.4e^2t + 0.2e^3t + 0.1e^4t

For (a) I said that it was a binomial distribution for part (i). Then for part (ii),
M'(t) = 5(0.3 + 0.7e^t)^4(0.7e^t)
M'(0) = 5(0.3 + 0.7e^0)^4(0.7e^0) = 5 (0.7) = 3.5

therefore ” = 3.5

M''(t) = 20(0.3 + 0.7e^t)^3(0.7e^t)^2 + 5(0.3 + 0.7e^t)^4(0.7e^t)
M''(0) = 20(0.3 + 0.7e^0)^3(0.7e^0)^2 + 5(0.3 + 0.7e^0)^4(0.7e^0) = 13.3

therefore σ^2 = M''(0) - [M'(0)]^2 = 13.3 - 3.5^2 = 1.05

For (b) I said that it was a Bernoulli distribution for part (i). Then for part (ii),
” = np = 1(0.55) = 0.55
σ^2 = np (1-p) = 0.55 ( 1- 0.55) = 0.2475

I don't understand your logic here ^^'
you use the formula np and np(1-p) for the Bernoulli, but not for the binomial ?
If you're asked to find mu and sigmaČ with the generating functions, then you should do the derivatives for Bernoulli too.
If you don't have to, then use the formula np and np(1-p) for the binomial (n is the power : 5). And p and p(1-p) for the Bernoulli

Quote:

For (c) I said that it was not a usual distribution
” = M'(0) = 2.1
True :)

Quote:

σ^2 = M''(0) - [M'(0)]^2 = 3.2 - 2.1 = 1.1 <---- I am very not sure here

Is my logic correct?
You forgot to square 2.1 !!! otherwise, it is correct ;)