M.g.f. help. Mean, Variance, and standard deviation.

[gimpy]

I am having trouble with this question.

Let X equal the number of flips of a fair coin that are required to observe the same face on consecutive flips.
(a) Find the p.m.f. of X.
if found the p.m.f. to be $$f(x) = (\frac{1}{2})^{x-1}$$ for $$x=2,3,4,...$$

(b) Give the values of the mean, variance and standard deviation of X.
For this one i found the m.g.f. to be $$M(t) = E(e^{tx}) = \sum_{x \in S} e^{tx}f(x)$$
$$M^{'}(0) = xf(x) = E(X)$$ which is the mean.
Then
$$M^{''}(0) = x^{2}f(x) = E(X)$$
$$Var(X) = M^{''}(0) - [M^{'}(0)]^2$$

Is this correct?

Than after that i realized that 2 rolls of the dice was the minimum to get the same face on two consecutive flips. So i made $$S={1,2}$$ and evaluted them like this getting the Mean = 2 and Variance = 4 which is not correct.
What am i doing wrong? Do i have to use Infinite series or something?

I haven't even started on the standard deviation.

(c) Find the values of
$$(i) P(X \leq 3)$$
$$(ii) P(X \geq 5)$$
$$(iii) P(X = 3)$$

I can't start on these until i get part (b)

 

[matt grime]

S must be the set {2,3,4...}. yes you need to evalaute the infinite sums, but they aren't hard to do. the sum of nr^n and n(n-1)r^n are quite well known and you should be able to find them.

you can do c without doing b; it doens't involve moments at all.


probability it happens on the first throw is 0, probability on the second is 1/2, and on the third is 1/4 so the first probability is 3/4

x greater than 5 is 1- prob of on the 2nd 3rd or 4th, which is 1-1/2-1/4-1/8

on the fifth exactly
well one of HTHTT or THTHH must happen and the probability is...?

 

[tacman]

figured out

First of all, great job on finding the p.m.f. for X. Your answer is correct.

For part (b), you are on the right track. The mean and variance can be found using the m.g.f. that you have calculated. However, there seems to be a mistake in your calculation of the variance. The correct formula for variance using the m.g.f. is Var(X) = M''(0) - [M'(0)]^2. So, in your case, Var(X) = M''(0) - [M'(0)]^2 = 2f(2) - [f(1)]^2 = 2(\frac{1}{2}) - (\frac{1}{2})^2 = \frac{3}{4}. Therefore, the variance is not 4 but \frac{3}{4}.

For the standard deviation, you just need to take the square root of the variance, so it would be \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.

For part (c), you can use the p.m.f. that you have found to calculate the probabilities.
(i) P(X \leq 3) = f(2) + f(3) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}
(ii) P(X \geq 5) = 1 - P(X \leq 4) = 1 - (f(2) + f(3) + f(4)) = 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8}) = \frac{1}{8}
(iii) P(X = 3) = f(3) = \frac{1}{4}

I hope this helps you understand the problem better. If you have any further questions, don't hesitate to ask for clarification. Good luck!