Moment generating function to calculate the mean and variance

1. Oct 27, 2012

snesnerd

I attached a pdf. The questions are not really what is stumping me. Its the wording of the question I dont understand. When it says, "Assume that 0 < λ < 1. Note that your answers will be in terms of the constant λ." and "Assume that λ > 0. Note that your answers will be in terms of the constant λ." I dont understand what they want me to really do here when it says that. To show I know how to do the questions before anyone helps me, I will briefly explain the solution to each one.

1a. Multiply by e^tx. Its a geometric series so just use the formula for the geometric series.
1b. To find the mean of X take the derivative of the answer to 1a.
1c. To find the variance of X take the second derivative minus first derivative squared.

2a. Multiply by e^tx then integrate from 0 to infinity.
2b. Same idea as 1b.
2c. Same idea as 1c.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Attached Files:

• probability.pdf
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2. Oct 27, 2012

Ray Vickson

I do not understand what your difficulty is. You have a distribution; it contains a parameter λ. You do not know the value of λ, but you are told it is between 0 and 1. The question asks you to determine the mean and variance. Of course there will be a λ in the answers, since there is a λ in the input. What is unclear about that?

RGV

3. Oct 27, 2012

HallsofIvy

Staff Emeritus
1a) asks you to find the moment generating function of the distribution $(1- \lambda)\lambda^x$ and you are told "1a. Multiply by $e^tx$." Okay, the moment generating function is, by definition, the expected value of $e^{tx}$ and the expected value of any function is the sum of function values for each x times the probabilty of that particular x. Here, that would be
$$\sum_{x=0}^\infty (1- \lambda)\lambda^xe^{tx}= (1- \lambda)\sum_{n=0}^\infty (\lambda e^t)^x$$
which is, as they say, a geometric series with common multiple $\lambda e^t$. Use the fact that the sum of a geometric series is given by
$$\sum_{x=0}^\infty r^x= \frac{1}{1- r}$$.

4. Oct 27, 2012

snesnerd

Thanks everyone. I dont know why I was letting it confuse me, but it seems clarified now. One last question though. What purpose did it serve to tell me that in the first question, 0 < λ < 1, and in the second question λ > 0? Couldnt they of just said that λ is just a constant?

5. Oct 27, 2012

Ray Vickson

No. In question 1 you need lambda between 0 and 1 in order to avoid negative probabilities or probabilities > 1. Question 2 is completely different; there you could have ANY positive value for lambda (but, again, a negative value would give nonsense such as negative probability, etc).

RGV

6. Oct 28, 2012

snesnerd

Gotcha. Thanks. Last question. The sum of the geometric series in the first question is

(1 - λ)/(1 - λe^t) right?

Since the first term is (1 - λ) and the common ratio is λe^t.