# Thread: k-th moment of X

1. ## k-th moment of X

With the probability density function:

$f(x)=\left\{\begin{array}{cc}\lambda e^{-\lambda x},&\mbox{ if } x>0;\lambda>0\\0, & \mbox otherwise \end{array}\right.$

i am asked to find the k-th moment of X, i.e. $E(X^k)$.

How do i do this?

2. You can do this by parts, but there's a nice trick to doing this in general for all gamma densities.

Let $X\sim \Gamma(\alpha, \beta)$

Then $f(x)={1\over \Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}I(x>0)$

And $E(X^k)={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}x^{k+\ alpha-1}e^{-x/\beta}dx$

$= {\Gamma(k+\alpha)\beta^{k+\alpha}\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}{1\ove r \Gamma(k+\alpha)\beta^{k+\alpha}}x^{k+\alpha-1}e^{-x/\beta}dx$

$= {\Gamma(k+\alpha)\beta^k\over \Gamma(\alpha)}$

Since that is a valid density in the integrand.
NOW you should know how the gamma function reduces (via parts)
and note that this is true even if k is not an integer.

Finally, let $\alpha=1$ and $\beta=1/\lambda$ and you have your answer.

3. As a follow-on it asks for the coefficient of skewness of X but doesn't give any $\alpha , \beta$ or $k$.

Is it possible to find an actual coefficient for this or does it just come out as a general solution?

4. I don't understand your question. In the end, you won't have anymore $\alpha,\beta$

baldeagle calculated the k-th moment of a gamma distribution with parameters $\alpha$ and $\beta$

Then just notice that if you substitute $\alpha=1, \beta=\frac 1\lambda$ in the pdf of a gamma distribution, you'll get the pdf you have.

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Another method... :

Let's calculate the moment generating function of your distribution.

\begin{aligned}M_X(t)&=\mathbb{E}(e^{tX}) \\
&=\lambda \int_0^\infty e^{tx}e^{-\lambda x} ~dx \\
&=\frac{\lambda}{\lambda-t} \end{aligned}

(defined for $t<\lambda$)

Then we know that the k-th moment of X will be the k-th derivative of its mgf, taken at t=0.

Find the Taylor series of $\frac{\lambda}{\lambda-t}$ when t is near 0 and its coefficients will be the derivatives.

$\frac{\lambda}{\lambda-t}=\frac{1}{1-\frac t\lambda}$ which is like a geometric series... :

$M_X(t)=\sum_{k=0}^\infty \left(\frac t\lambda\right)^k=\sum_{k=0}^\infty t^k \cdot \frac{k!}{\lambda^k}\cdot \frac{1}{k!}$

But the Taylor series of $M_X(t)$ is $\sum_{k=0}^\infty t^k \cdot \frac{M^{(k)}(0)}{k!}$
where $M^{(k)}(0)$ denotes the k-th derivative of M taken at t=0.

By identification, it follows that $\mathbb{E}(X^k)=M^{(k)}(0)=\frac{k!}{\lambda^k}$

5. But, be aware I calculated any moment as long as $k>-\alpha$
That include noninteger and negative moments.
Using the MGF you are only getting the positive integer moments.

6. And since $\alpha=1$, it's like calculating for $k\geq 0$... So it's not about negative moments.
And how can one define a negative moment ? oO

7. Originally Posted by Moo
And since $\alpha=1$, it's like calculating for $k\geq 0$... So it's not about negative moments.
And how can one define a negative moment ? oO

No, for k>-1, in that case. Of course one can find $E(X^{-1/2})$

8. Ah, sorry, the first time i saw the page the bottom part about $\alpha=1, \beta=\frac{1}{\lambda}$ didn't appear.