Find (without using MGF) the mean and variance.

$$f(x) = \exp(-kx)x^{(r-1)}k^r/(r-1)!\ \text{ for }\ x>=0$$

$$f(x) = 0\ \text{ for }\ x<0$$

$r$ positive integer, $k>0$

  • $\begingroup$ Looks like this should be tagged homework. What have you tried? $\endgroup$
    – Glen_b
    Dec 22 '12 at 4:44
  • 1
    $\begingroup$ Hint: integration by parts. $\endgroup$ Dec 22 '12 at 13:18
  • 1
    $\begingroup$ Hint 2: What is the constant of integration? How would that change if you put an extra $x$ in front, as when integrating to find the mean? $\endgroup$
    – jbowman
    Dec 22 '12 at 16:25
  • $\begingroup$ Tried by parts but it doesn't work. The answers are as follows but I cannot seem to derive it. Mean = r/k, Variance = r/k^2 $\endgroup$
    – Sharingan
    Dec 22 '12 at 17:18

The answer is actually already there in the question, staring us in the face. All we need to do is interpret it appropriately.

Look at the constants in the definition of $f$: $k^r$ divided by $(r-1)!$. The first factor should really be absorbed by the $x$'s and the implicit $dx$. That is,

$$k^r x^{r-1} dx = (kx)^{r-1} d(kx).$$

This makes it clear that the distribution is really a function of $kx$, revealing $k$ as a scale factor. If we can find the mean $\mu$ and variance $\sigma^2$ for $k=1$, we would only need to divide them by $k$ and $k^2$, respectively, to get the fully general answer.

That observation reduces the question to one of finding various moments of

$$f_r(x) = \frac{x^r \exp(-x)}{x (r-1)!}.$$

There's only one constant in there, and it must be there to make sure this distribution integrates to unity. (That is the import of @jbowman's hint in the comments.) In other words, you already know that

$$\int_0^\infty x^r \exp(-x) \frac{dx}{x} = (r-1)!.$$

To find the $j$th moment ($j=1,2$ are all that are needed), you must obtain a formula for

$$\int_0^\infty x^j f_r(x) dx = \int_0^\infty x^j\left(x^r \exp(-x) \frac{dx}{x}\right) = \int_0^\infty x^{r+j} \exp(-x) \frac{dx}{x}.$$

It is now but a moment's work to apply the penultimate formula to this expression to obtain the answers. I leave the details--which are simple algebra, since the integration has been done--to the interested reader.

  • $\begingroup$ The somewhat strange-looking notation $\frac{dx}{x}$ is employed because the integral can be understood as occurring over the multiplicative group of positive real numbers and this is its Haar measure (the unique invariant measure, up to a constant multiple). It's also a neat way to connect these integrals with the Gamma function with parameter $r$ (not $r-1$, as it would seem from the conventional notation). $\endgroup$
    – whuber
    Dec 22 '12 at 18:30
  • $\begingroup$ Upvoted because of the Haar mesure! $\endgroup$
    – Elvis
    Dec 24 '12 at 6:53
  • $\begingroup$ Thanks. Is it possible to use the fact that the integral of a pdf has probability 1? $\endgroup$
    – Sharingan
    Dec 26 '12 at 20:45
  • $\begingroup$ Yes: that is exactly what the penultimate equation in my answer is saying. The whole point is that you can substitute anything (except zero or a negative integer) for $r$ in that equation. By substituting $r+j$, you immediately get a formula for the last equation. $\endgroup$
    – whuber
    Dec 26 '12 at 20:50

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