- #1
Saladsamurai
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I am using a particular form of the incomplete gamma function (which I have never seen before) in my probability course. It is denoted:
[tex]F(x;\alpha) = \int_0^x \frac{y^{\alpha - 1}e^{-y}}{\Gamma(\alpha)}\,dy\qquad(1)[/tex]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,
[tex] P(X\le x)=\int_0^x \frac {1}{\beta^\alpha\Gamma(\alpha)}x^{\alpha - 1}e^{-x/\beta}\,dx\qquad(2)[/tex]
using [itex]y = x/\beta[/itex] (3), but I am having a little trouble getting there. Obviously using (3), the [itex]e^-{x/\beta}[/itex] term becomes the desired e^-y as in (1). If [itex]y = x/\beta[/itex], then [itex]x = y*\beta[/itex]. So I am left with,
[tex] P(X\le x)=\int_0^x \frac {1}{\beta^\alpha\Gamma(\alpha)}(y\beta)^{\alpha - 1}e^{-y}\,dy\qquad(4)[/tex]
[tex]\Rightarrow P(X\le x)=\int_0^x \frac {y^{\alpha - 1}\beta^{\alpha - 1}e^{-y}}{\beta^\alpha\Gamma(\alpha)}\,dy\qquad(5)[/tex]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 [itex]1/\beta[/itex] can be pulled through the integral, but the text solution to a particular problem does not have that factor of [itex]1/\beta[/itex] left in there, hence I think I am missing something.)
[tex]F(x;\alpha) = \int_0^x \frac{y^{\alpha - 1}e^{-y}}{\Gamma(\alpha)}\,dy\qquad(1)[/tex]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,
[tex] P(X\le x)=\int_0^x \frac {1}{\beta^\alpha\Gamma(\alpha)}x^{\alpha - 1}e^{-x/\beta}\,dx\qquad(2)[/tex]
using [itex]y = x/\beta[/itex] (3), but I am having a little trouble getting there. Obviously using (3), the [itex]e^-{x/\beta}[/itex] term becomes the desired e^-y as in (1). If [itex]y = x/\beta[/itex], then [itex]x = y*\beta[/itex]. So I am left with,
[tex] P(X\le x)=\int_0^x \frac {1}{\beta^\alpha\Gamma(\alpha)}(y\beta)^{\alpha - 1}e^{-y}\,dy\qquad(4)[/tex]
[tex]\Rightarrow P(X\le x)=\int_0^x \frac {y^{\alpha - 1}\beta^{\alpha - 1}e^{-y}}{\beta^\alpha\Gamma(\alpha)}\,dy\qquad(5)[/tex]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 [itex]1/\beta[/itex] can be pulled through the integral, but the text solution to a particular problem does not have that factor of [itex]1/\beta[/itex] left in there, hence I think I am missing something.)
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