Questions tagged [moment-generating-functions]

For questions relating to moment-generating-functions (m.g.f.), which are a way to find moments like the mean$~(μ)~$ and the variance$~(σ^2)~$. Finding an m.g.f. for a discrete random variable involves summation; for continuous random variables, calculus is used.

A moment generating function (MGF) is a single expected value function whose derivatives produce each of the required moments.

Definition: Let $X$ be a discrete random variable with probability mass function $f(x)$ and support $S$. Then:

$$M_X(t) = E(e^{tX})=\sum\limits_{x\in S} e^{tx}f(x)$$or, $$M_X(t) = E(e^{tX}) = \int_x e^{tx} f(x) \, \mathrm{d}x$$

is the MGF of $X$ as long as the summation is finite for some interval of $t$ around $0$.

i.e. $M(t)$ is the MGF of $X$ if there is a positive number $h$ such that the above summation exists and is finite for $−h<t<h$.

Note: There are basically two reasons for which MGF's are so important.

  • the MGF of $X$ gives us all moments of $X$.
  • the MGF (if it exists) uniquely determines the distribution. That is, if two random variables have the same MGF, then they must have the same distribution.

Thu if you find the MGF of a random variable, you have indeed determined its distribution.

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Obtaining the probability density function from the moment generating function problem

I'm trying to find a formula to find the pdf from the distribution's moments. I did the following: $$ f_X(x) = \int_{0}^{\infty} M_X(t)\text{e}^{-tx} dt = \int_{0}^{\infty} \sum_{k=0}^{\infty} \frac{t^k}{k!} E \left[ X^k\right]\text{e}^{-tx} dt = …
Felipe
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moment generating functions of positive random variables

If the moment generating function $M_X(\cdot)$ of a random variable is known, is there a way to tell whether $P(X\geq 0) = 1$?
jochen
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Opposite of 'If X,Y are independent, Mx+y(t)=Mx(t)My(t)'

It is known that If X,Y are independent random variables, Mx+y(t)=Mx(t)My(t) It's because if X and Y are independent, f(X) and g(Y) are also independent. How about the opposite? It X and Y have moment generating function and Mx+y(t) = Mx(t)My(t),…
Damelim
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moment generating function of the difference between two random variables

I need to find the moment generating function of $G = Y - X$ where $Y$ ~ exp($\frac{1}{2}$) and $X$ ~ exp(1). X and Y are independent. I read in another topic that $m_{Y-X}(r) = \frac{m_Y(r)}{m_X(r)}$ but I couldn't find a proof. I started as…
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How to find the MGF of this pmf?

A random variable $X$ has pmf $p(x;\alpha) = (1-\alpha)^{x-1} \alpha$ for $x = 1,2,\dots$. Find the moment generating function of $X$, $M_X(t)$. What I've done: $E[e^{tx}] = \sum_x e^{tx} (1-\alpha)^{x-1} \alpha = \frac{\alpha}{1-\alpha} \sum_x (e^t…
goblinb
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PDF for Moment Generating Function $(1 +\beta t)^{-\alpha}$ when $\beta > 0$

I'm trying to find the PDF for the following MGF. $$(1 + t/2)^{-3/2}$$ I already know that $(1 - \beta t)^{-\alpha}$ when $\beta > 0$ is an MGF for the Gamma distribution $\Gamma(\alpha, \beta$). However, the above formula since it is clearly not…
Moreblue
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With K(t) = log(E[e^tX]) ,show that K'(0) = E[X]

Just as the title, it is Question 83 on page 90(Ross's book, Introduction to Probability Models) since K(t) = $log(E[e^{tX}])$ Convert $log_{10}$ to $ln_e$, K(t) = $ln(E[e^{tX}])$/$ln$10 Then K'(t) = $E[Xe^{tX}]$ / ($E[e^{tX}]$ * $ln$10 ) K'(0) =…
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What is cumulative generating function of this probability measure?

I am trying to find what is CGF of this probability measure: $$\mu=\alpha \delta_a+(1-\alpha) \delta_b$$ I found it difficult because when I tried to calculate Moment generating function, I didn't know what is $\mu(dx)$ (which is density function)…
Melina
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Find moment generating function of another function

fx(x) = 2x/x^2 is a pdf of random variable X 0 < x < c Then Fx(x) = x^2/c^2 Find moment generating function of Y Y = 2x+10, note it is increasing. X = (Y-10)/2 lets call this function g(y) Fy(y)= Fx(g(y)) = 1/(c^2)*((y-10)/2))^2 No? And when Fy(y)…
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MGF of quadratically transformed normal rv

For Z~N(1,1/2), find the moment generating function of $W=Z^2$. $$f_z(z)= \frac{1}{\sqrt{2\pi\sigma^2}}e^\frac{-(z-\mu)^2}{2\sigma^2}=\frac{1}{\sqrt\pi}e^{-(z-1)^2}$$ so $$M_W(s)=E[e^{sW}]=E[e^{sZ^2}]$$ but I can't get anywhere trying to evaluate…
Padster
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What is the distribution ?

Q=================================================================== $f(Θ)$ is pdf of gamma distribution $$f(Θ) = \frac{λ^α}{Γ(α)}Θ^{α-1}\exp(-λΘ), $$ $$X\mid Θ \sim \mathrm{poisson}(Θ) \rightarrow \frac{Θ^x\exp(-Θ)}{x!}$$ Suppose that $Θ$ is a…
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Using the mgf of a distribution to find expected values of an estimator

Suppose that $Y_1, ..., Y_n$ is a random sample from a Poisson distribution with unknown mean $\theta$ and let $T = \sum_{i = 1}^n Y_i$. Find the constant $c$ such that the estimator $e^{-cT}$ is unbiased as an estimator of $e^{-\theta}$. To show…
Adrian
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Figuring out MFG

Let X have a Poisson distribution with parameter L(short for lambda here ) show that the MGF function of Y=(x-L)/sqrt(L) is given by m(t) = exp(lamda*e^(t/sqrt(lamda)) - sqrt(lamda)t - lamda) Here's what I did: My(t)=E(e^Yt) …
user1919987
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Moment generating function $Y$

If $X$ is a random variable, normally distributed with unknown parameters how could I find the mgf of random variable $Y$, where Y=$e^X$? I am able to find mgf of $X$ from the mgf of a standard normally distributed random variable but not for $Y$.…
Rivaldo
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