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MGF of this

nacho

Active member
Sep 10, 2013
156
Please refer to the attached image.


The concept of MGF still plagues me.

I got an invalid answer when i tried this.

What i did was:

$ \int e^{tx}f_{X}(x)dx $
= $ \int_{-\infty}^{+\infty} e^{tx}(p \lambda e^{-\lambda x} + (1-p)\mu e^{-x\mu})dx$

I was a bit wary at this point, because it reminded me of the bernoulli with the p and (1-p) but i could not find any relation for this.

i separated the two integrals, and ended up with
$ p \lambda \int_{-\infty}^{+\infty}e^{tx-x\lambda}dx + ... $ which i knew was immediately wrong because that integral does not converge.
What did i do wrong.

What does the MGF even tell us. First, second, nth moment, what does this mean to me?
 

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chisigma

Well-known member
Feb 13, 2012
1,704
Please refer to the attached image.


The concept of MGF still plagues me.

I got an invalid answer when i tried this.

What i did was:

$ \int e^{tx}f_{X}(x)dx $
= $ \int_{-\infty}^{+\infty} e^{tx}(p \lambda e^{-\lambda x} + (1-p)\mu e^{-x\mu})dx$

I was a bit wary at this point, because it reminded me of the bernoulli with the p and (1-p) but i could not find any relation for this.

i separated the two integrals, and ended up with
$ p \lambda \int_{-\infty}^{+\infty}e^{tx-x\lambda}dx + ... $ which i knew was immediately wrong because that integral does not converge.
What did i do wrong.

What does the MGF even tell us. First, second, nth moment, what does this mean to me?
By definition is...

$\displaystyle M(t) = E \{ e^{t\ X} \} = \int_{- \infty}^{+ \infty} f(x)\ e^{t\ x}\ dx = \int_{0}^{\infty} \{p\ \lambda\ e^{- \lambda\ x} + (1-p)\ \mu\ e^{- \mu\ x}\ \}\ e^{t\ x}\ d x = \frac{p}{1 - \frac{t}{\lambda}} + \frac{1-p}{1-\frac{t}{\mu}}\ (1)$

The knowledge of M(t) permit us to find mean and variance of X with the formula...

$\displaystyle E \{X^{n}\} = M^{(n)} (0)\ (2)$

... so that is...

$\displaystyle E \{X\} = \frac{p}{\lambda} + \frac{1-p}{\mu}\ (2)$

$\displaystyle E \{X^{2}\} = \frac{2\ p}{\lambda^{2}} + \frac{2\ (1-p)}{\mu^{2}}\ (3)$

$\displaystyle \sigma^{2} = E \{X^{2} \} - E^{2} \{ X \} = \frac{2\ p - p^{2}}{\lambda^{2}} + \frac{2\ (1-p) - (1-p)^{2}}{\mu^{2}} - 2\ \frac{p\ (1-p)}{\lambda\ \mu}\ (4)$

Kind regards

$\chi$ $\sigma$
 

nacho

Active member
Sep 10, 2013
156
By definition is...

$\displaystyle M(t) = E \{ e^{t\ X} \} = \int_{- \infty}^{+ \infty} f(x)\ e^{t\ x}\ dx = \int_{0}^{\infty} \{p\ \lambda\ e^{- \lambda\ x} + (1-p)\ \mu\ e^{- \mu\ x}\ \}\ e^{t\ x}\ d x = \frac{p}{1 - \frac{t}{\lambda}} + \frac{1-p}{1-\frac{t}{\mu}}\ (1)$

$\chi$ $\sigma$

I don't see how this integral converges, how did you get that answer
 

chisigma

Well-known member
Feb 13, 2012
1,704
I don't see how this integral converges, how did you get that answer
Is...

$\displaystyle \lambda\ \int_{0}^{\infty} e^{- (\lambda-t)\ x}\ d x = \frac{\lambda}{t - \lambda} |e^{- (\lambda-t)\ x}|_{0}^{\infty} = \frac{1}{1-\frac{t}{\lambda}}\ (1)$

... and [of course...] the integral in (1) converges if $\displaystyle t< \lambda$. That is not a disavantage because from the pratical point of view what matters in the behaviour of M(t) in t=0...

Kind regards

$\chi$ $\sigma$