Unravelling the Mystery of MGFs: Exploring Moments and More

[nacho-man]

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?

 

[chisigma]

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-man]

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]

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

$\chi$ $\sigma$

 

[blue_raver22]

As a scientist, it is important to approach problems with a critical and analytical mindset. In this case, the issue seems to be with the integration of the MGF formula. It is important to carefully check the steps and make sure all assumptions and conditions are met before plugging in values.

Additionally, it is important to understand the concept of MGF and its applications. The MGF, or moment generating function, is a powerful tool in probability and statistics that allows us to calculate moments of a random variable. The moments, such as the first, second, and nth moment, provide important information about the distribution of the random variable.

In this case, it appears that the MGF formula was not properly applied, leading to an invalid answer. It is important to carefully review the formula and make sure all variables and constants are correctly identified and accounted for.

Furthermore, it is important to understand the relationship between the MGF and other probability distributions, such as the Bernoulli distribution. This can provide insight into how the MGF can be used to calculate moments and other important properties of a random variable.

In conclusion, the MGF can be a complex concept, but with careful application and understanding, it can provide valuable insights and help unravel mysteries in the world of probability and statistics.