# Homework Help: Incomplete Gamma Function Notation

1. Jun 20, 2010

I am using a particular form of the incomplete gamma function (which I have never seen before) in my probability course. It is denoted:

$$F(x;\alpha) = \int_0^x \frac{y^{\alpha - 1}e^{-y}}{\Gamma(\alpha)}\,dy\qquad(1)$$

Question 1
Why the bounds in terms of 'x' ? I am just a little confused by this. Is 'x' just being used as a dummy variable? That is 'x' is some particular value of 'y' ? I know, stupid question, but I don't want to make a bad assumption here.

Question 2
My text has values of the integral in (1) tabulated. I am supposed to use (1) to solve the integral,

$$P(X\le x)=\int_0^x \frac {1}{\beta^\alpha\Gamma(\alpha)}x^{\alpha - 1}e^{-x/\beta}\,dx\qquad(2)$$

using $y = x/\beta$ (3), but I am having a little trouble getting there. Obviously using (3), the $e^-{x/\beta}$ term becomes the desired e^-y as in (1). If $y = x/\beta$, then $x = y*\beta$. So I am left with,

$$P(X\le x)=\int_0^x \frac {1}{\beta^\alpha\Gamma(\alpha)}(y\beta)^{\alpha - 1}e^{-y}\,dy\qquad(4)$$

$$\Rightarrow P(X\le x)=\int_0^x \frac {y^{\alpha - 1}\beta^{\alpha - 1}e^{-y}}{\beta^\alpha\Gamma(\alpha)}\,dy\qquad(5)$$

My trouble now is that the order of the Beta in the numerator is one less than that in the denominator. So I am left with a Beta in the denominator. Did I miss something or have I erred? It "almost" looks like (1), but not quite. (I know that a constant of $1/\beta$ can be pulled through the integral, but the text solution to a particular problem does not have that factor of $1/\beta$ left in there, hence I think I am missing something.)

Last edited: Jun 20, 2010
2. Jun 20, 2010

### Staff: Mentor

No, the dummy variable is y. The expression on the right side is a function of x as well as the parameter alpha. Your second question is a little more complicated, so I'll need to think about it some more.

3. Jun 20, 2010

Hmmm... How can a function F(x;a) be given by an integral over 'y' ? Sorry. I was never good at the formal stuff.

4. Jun 20, 2010

### Staff: Mentor

Just a quick thought that might or might not solve your problem. If x = $\beta$y, then dx = $\beta$dy. I'm not sure that you accounted for this.

5. Jun 20, 2010

### Staff: Mentor

Let's look at a simpler example. Consider this function of x:
$$f(x) = \int_0^x t^2 dt$$
Clearly f(0) = 0. What is f(1)? f(2)? In this integral t is a dummy variable; it could be y or v or whatever (just not x). The real variable here is the upper limit of integration.

6. Jun 20, 2010

Oh bloody hell! (What am I British now? Why would I say bloody hell?) That is it! Thanks!

Ok. This reminds me of the notation on one of the fundamental laws of calculus, which 3 years later I still do not understand. I will look at what you have written and figure it out. So what is f(1)? Isn't it

$$\int_0^1 t^2\,dt$$

So what we are saying is that that f(x) is a function of the upper limit of integration, correct? As opposed to what is under the integrand. Err...maybe not....

7. Jun 20, 2010

### Staff: Mentor

That's pretty much what I'm saying, although it's not limited to just the upper limit of integration. The following are also functions of x:
$$f(x) = \int_x^0 v^2 dv$$
$$g(x) = \int_x^{2x} r^2 dr$$
Note that the integrand is the same as before, but I have changed the dummy variables. Of course, the integrand plays a role in what you get, but it doesn't matter whether the integrand is written in terms of t or v or whatever, so it is called a dummy variable.

8. Jun 20, 2010

OK. But here is another point of confusion. This is the problem from which this discussion stems. It is a particular example that I was trying to generalize (since the text does not). Sorry for the crappy image, my scanner just crapped out on me

Notice that the first integral is the gamma distribution given by (2) in the OP. The third integral is the 'converted' integral as given by (1) in the OP. But clearly the bounds on the third integral have been converted to 'y' bounds to match the function's dummy variable using 'y = x/beta'. Hence they run from 0 to 6 instead of 0 to 60.

Is this simply because if we left the bounds of the new integral in terms of x, it would not be of any use to us; i.e. we could not evaluate it? But since we know the relationship beween x and the dummy variable, we can substitute it in?

9. Jun 20, 2010

### Staff: Mentor

There's a difference between the second and third integrals on your scanned page: the second one is a function of x, while the third is not a function at all - it's just a number. In the 2nd integral, from P(X <= 60), it's evident that x = 60. Making the change from x to y (with y = x/beta), you have y = 60/10 = 6. That's the reason for the change in the upper limit of integration. For the lower limit, you have y = 0/10 = 0, so it doesn't change.

10. Jun 20, 2010

I am sorry Mark, but you are kind of losing me. In the scanned document, the second integral and the third are the same thing. The probability the X <= x is given by evaluating the 2nd integral from 0 to x. That is,

$$P(X\le x) = F(x;\alpha) = \int_0^x \frac{y^{\alpha-1}e^{-y}}{\Gamma(\alpha)}\, dy$$

For this problem, x takes on the value of 60, beta = 10 and alpha = 5.

So we need to evaluate

$$F(60;5) = \int_{x=0}^{x=60} \frac{y^{4}e^{-y}}{\Gamma(5)}\, dy$$

But the integral is wrt 'y' and our bounds are in x, so we must convert to y bounds using the fact that x we defined our dummy variable 'y' interms of 'x'. That is all I am saying. Are you just pointing out that once we do that, we no longer have a function since it evaluates to a numerical value?

11. Jun 20, 2010

### Staff: Mentor

Not at all. The 2nd integral is a function of x (and also of a parameter alpha). The 3rd integral is a number. The 3rd integral is certainly related to the 2nd integral by virtue that it represents f(60). A function f(x) and a number f(60) are very different things.
Yes.

12. Jun 20, 2010