# Mistake in calculating characteristic function of the gamma distribution

I am trying to partially prove that the characteristic funtion of a Gamma$$(\alpha,\lambda)$$ distribution, $$\varphi(t)$$, is such that $$\varphi(t)=\left(\frac\lambda{\lambda-it}\right)^\alpha.$$ I am only trying to prove this for $$|t|<\lambda$$. However, my workings have led me to a different answer and I can't see what I have done wrong.

We have $$\varphi(t)=\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}e^{itx}dx=\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\sum_{n=0}^\infty\frac{(itx)^n}{n!}dx.$$ Now I want to try to swap the integral and the sum, so I will try to apply Fubini's theorem. Consider \begin{align*} \sum_{n=0}^\infty\int_0^\infty\left|\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(itx)^n}{n!}\right|dx=&\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(|t|x)^n}{n!}dx\\ =&\sum_{n=0}^\infty\left(\frac{|t|}\lambda\right)^n\int_0^\infty\frac{\lambda^{\alpha+n}}{\Gamma(\alpha+n)}x^{\alpha+n-1}e^{-\lambda x}dx\\ =&\sum_{n=0}^\infty\left(\frac{|t|}\lambda\right)^n\\ <&\infty \end{align*} The third equality comes from comparing the integral to a Gamma$$(\alpha+n,\lambda)$$ density, and the fourth line (the inequality) comes from the fact that I assumed $$|t|<\lambda$$. So now we can swap the integral and the sum. So \begin{align*} \varphi(t)&=\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(itx)^n}{n!}dx\\ &=\sum_{n=0}^\infty\left(\frac{it}{\lambda}\right)^n\int_0^\infty\frac{\lambda^{\alpha+n}}{\Gamma(\alpha+n)}x^{\alpha+n-1}e^{-\lambda x}dx\\ &=\sum_{n=0}^\infty\left(\frac{it}\lambda\right)^n\\ &=\frac\lambda{\lambda-it}. \end{align*}\\ However, this is different to what I know the answer obviously should be, but I cannot see where I went wrong. Is this a valid approach? What should I do instead? Any help appreciated!

• $\Gamma (\alpha) n!$ is not equal to $\Gamma (\alpha+n)$. Aug 4, 2021 at 9:41
• Oh yeahh thanks haha! Should I delete this question? Aug 4, 2021 at 9:43
• You can keep it. See this for a proof: statlect.com/probability-distributions/gamma-distribution Aug 4, 2021 at 9:46

Consider \begin{align*} &\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(|t|x)^n}{n!}dx\\ =&\sum_{n=0}\frac{\Gamma(\alpha+n)}{\Gamma(\alpha)n!}\left(\frac{|t|}\lambda\right)^n\int_0^\infty\frac{\lambda^{\alpha+n}}{\Gamma(\alpha+n)}x^{\alpha+n-1}e^{-\lambda x}dx\\ =&\sum_{n=0}^\infty{\alpha+n-1\choose n}\left(\frac{|t|}\lambda\right)^n\\ =&\sum_{n=0}^\infty(-1)^n{-\alpha\choose n}\left(\frac{|t|}\lambda\right)^n\\ =&\sum_{n=0}^\infty{-\alpha\choose n}\left(-\frac{|t|}\lambda\right)^n\\ =&\left(1-\frac{|t|}\lambda\right)^{-\alpha}<\infty. \end{align*} So we can swap the sum and the integral. And now we can apply a smiliar argument to get $$\varphi(t)=\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(itx)^n}{n!}dx=\left(\frac{\lambda-it}\lambda\right)^{-\alpha}.$$ Edit: The only thing I am now unsure of is when did I use that $$|t|<\lambda$$?