1. ## Discrete Distributions

(i)Given the name of the distribution of X; (ii)find the values of µ 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 µ = 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),
µ = 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
µ = 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?

2. Originally Posted by MathRules!
(i)Given the name of the distribution of X; (ii)find the values of µ 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 µ = 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),
µ = 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
µ = 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
$M(t) = (0.3 + 0.7e^t)^5$ 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.
$M_X(t)=\sum_xP(X=x)e^{xt}=.3e^t + .4e^{2t} + .2e^{3t} +.1e^{4t}$
Now just match up terms. The coefficients are the probabilities and that's why they sum to one, and $M(0)=1$.
And the constants in front of the t in the exponent are just the realizations of x.

Thus $P(X=1)=.3,P(X=2)=.4, P(X=3)=.2,P(X=4)=.1$.

3. Hello,
Originally Posted by MathRules!
(i)Given the name of the distribution of X; (ii)find the values of µ 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 µ = 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),
µ = 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 sigma² 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

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

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