Puzzle with moment generating function for gamma function

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Homework Help Overview

The discussion revolves around the moment-generating function (mgf) for a random variable, specifically in the context of the gamma distribution and its relationship to the exponential distribution. Participants are examining the moments of these distributions and how they relate to the provided formulas.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to compute the moments of a gamma-distributed random variable and are comparing these to the moments of an exponential distribution. There are questions about the definitions and computations of the moment-generating function, as well as the implications of assuming relationships between the two distributions.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the definitions and computations related to the moment-generating function. Some have provided detailed steps in their calculations, while others are questioning the assumptions made regarding the distributions involved. There is a recognition of the need for more detailed information to assist in resolving the confusion.

Contextual Notes

There is a noted confusion regarding the characterization of the random variable as either exponential or gamma, which has implications for the moment calculations. Participants are also discussing the significance of specific parameters and their effects on the moments derived from the distributions.

Askhwhelp
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I am given that The kth moment of an exponential random variable with mean mu is
E[Y^k] = k!*mu^k for nonnegative integer k.

I found m^2 (0) = (-a)(-a-1)(-beta)^2. The answer I found is however mu^2+a*beta^2 which is different from the k! From the given formula.
Could someone help me figure it out what happen?
 
Last edited:
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Askhwhelp said:
I am given that The kth moment of an exponential random variable with mean mu is
E[Y^k] = k!*mu^k for nonnegative integer k.

I found m^2 (0) = (-a)(-a-1)(-beta)^2. The answer I found is however mu^2+a*beta^2 which is different from the k! From the given formula.
Could someone help me figure it out what happen?

You need to show all the steps, in detail, because you are certainly obtaining the wrong answer.
 
Ray Vickson said:
You need to show all the steps, in detail, because you are certainly obtaining the wrong answer.
m(t) = (1-beta*t)^-a

m'(t) = -a * (1-beta*t)^(-a-1)*(-beta)
m'(0) = -a *(-beta)
m''(t) = -a*(-a-1)* (1-beta*t)^(-a-1)*(-beta)^2
m''(0) = -a*(-a-1)* (-beta)^2
m'''(t) = -a*(-a-1)*(-a-2)*(1-beta*t)^(-a-1)*(-beta)^3
m'''(0) = a*(-a-1)*(-a-2)*(-beta)^3
 
Askhwhelp said:
m(t) = (1-beta*t)^-a

m'(t) = -a * (1-beta*t)^(-a-1)*(-beta)
m'(0) = -a *(-beta)
m''(t) = -a*(-a-1)* (1-beta*t)^(-a-1)*(-beta)^2
m''(0) = -a*(-a-1)* (-beta)^2
m'''(t) = -a*(-a-1)*(-a-2)*(1-beta*t)^(-a-1)*(-beta)^3
m'''(0) = a*(-a-1)*(-a-2)*(-beta)^3

You are supplying no useful information at all. I need details, right from the very beginning. In other words, I need to know: (1) What is your formula for the density function of Y? (2) What is the definition of the moment-generating function m(t)? (3) For your particular f, what is the integration you perform to get your m(t)?

Only after I see all that can I possibly help you. I know the definition of m(t), but the question is whether YOU know it. Judging from your formula for m(t), I would guess the answer to be NO, you do not know it. I need to see that. I need details, details, details!
 
Ray Vickson said:
You are supplying no useful information at all. I need details, right from the very beginning. In other words, I need to know: (1) What is your formula for the density function of Y? (2) What is the definition of the moment-generating function m(t)? (3) For your particular f, what is the integration you perform to get your m(t)?

Only after I see all that can I possibly help you. I know the definition of m(t), but the question is whether YOU know it. Judging from your formula for m(t), I would guess the answer to be NO, you do not know it. I need to see that. I need details, details, details!

f(y) = y^(a-1)*e^(-y/beta)/(beta^(a)*Γ(a))

moment-generating function m(t) is alternative specification for its probabilit function, E(Y^k) = m^k (0)
 
Askhwhelp said:
f(y) = y^(a-1)*e^(-y/beta)/(beta^(a)*Γ(a))

moment-generating function m(t) is alternative specification for its probabilit function, E(Y^k) = m^k (0)

Your first message said that Y had an exponential distribution (or so it seemed so say). Is that not true? Does Y really have a (general) Gamma distribution? I will assume so, from what you have now just written. Do you see why your messages cause confusion?

OK, so now tell me: what is the definition of m(t)? Don't just tell me "... m(t) is alternative specification for its probabilit function..."; tell me what it IS! Then I will be able to judge whether you have a correct formula or not.
 
Ray Vickson said:
Your first message said that Y had an exponential distribution (or so it seemed so say). Is that not true? Does Y really have a (general) Gamma distribution? I will assume so, from what you have now just written. Do you see why your messages cause confusion?

OK, so now tell me: what is the definition of m(t)? Don't just tell me "... m(t) is alternative specification for its probabilit function..."; tell me what it IS! Then I will be able to judge whether you have a correct formula or not.


I think My first message is that gamma distribution while exponential distribution is (1/beta)*e^(-t/beta). What makes you think my message causes confusion?
m(t) = E(e^(tY))
 
Askhwhelp said:
I think My first message is that gamma distribution while exponential distribution is (1/beta)*e^(-t/beta). What makes you think my message causes confusion?
m(t) = E(e^(tY))

Because it confused me. Here is EXACTLY what you said:

"I am given that The kth moment of an exponential random variable with mean mu is E[Y^k] = k!*mu^k for nonnegative integer k.
I found m^2 (0) = (-a)(-a-1)(-beta)^2. The answer I found is however mu^2+a*beta^2 which is different from the k! From the given formula.
Could someone help me figure it out what happen?"

You did not mention the Gamma distribution anywhere in that first message---only the exponential!

Anyway, your computation of the mgf for Gamma-distributed Y is OK, and your subsequent computations of ##m^{(k)}(0)## are OK for k = 1 and 2, but your ##m'''(0)## has the wrong sign. You should always simplify carefully before substitution k=0, and that means you should always get rid of the negatives in factors such as (-a), etc---take out the '-' signs carefully.

As to your original question: why would you assume the moments of the exponential distribution have anything to do with the moments of the Gamma distribution?
 
Last edited:
Ray Vickson said:
"As to your original question: why would you assume the moments of the exponential distribution have anything to do with the moments of the Gamma distribution?

Let holds this a little bit. I need to figure out what context I took this from
 
  • #10
Askhwhelp said:
Let holds this a little bit. I need to figure out what context I took this from

Note that the exponential is a special case of the gamma, obtained by setting b = 1. What happens to b(b+1)(b+2) ... (b+k-1) when you put b = 1?
 

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