I try to calculate expected value of exponential distribution with integration by parts without success. I know that the text book way to do it so that first take lambda out of the integral and then do the integration by parts. What if I would like to calculate it directly: \begin{align} E[X]=\int_{0}^\infty x\lambda e^{-\lambda x} dx \end{align} And now \begin{align} f(x)=x, \quad f'(x)=dx \end{align} and \begin{align} g'(x)=-\lambda e^{-\lambda x}dx, \quad g(x)= e^{-\lambda x} \end{align} So I would like to start with \begin{align} E[X]=-\int_{0}^\infty x (-\lambda e^{-\lambda x}) dx. \end{align} Is this possible and if not why not?

  • $\begingroup$ I mean sure, but its exactly what you started with. You haven't made any real progress. $\endgroup$
    – K.defaoite
    Apr 16 at 14:02
  • 1
    $\begingroup$ Choose $u(x)=x$ and $v^{'}(x)=-\lambda e^{-\lambda x}$. Then it works. Advice: Take care of the signs (+,-). $\endgroup$ Apr 16 at 14:05
  • $\begingroup$ I think this is exactly what I try to do and the question is how it works. $\endgroup$
    – Parallax
    Apr 16 at 14:06
  • $\begingroup$ @Parallax Give a reply whether you succeeded or not. $\endgroup$ Apr 16 at 14:14
  • $\begingroup$ I didn't. I will add my calculations to my question. $\endgroup$
    – Parallax
    Apr 16 at 14:21

In this case it is much easier to use the fact (which follows from Tonelli's theorem) that for a non-negative random variable $X$ with finite mean, $\mathbb E[X] = \int_0^\infty (1-F_X(x))\ \mathsf dx$, with $F_X$ being the distribution function of $X$. Hence $$ \mathbb E[X] = \int_0^ \infty e^{-\lambda t}\ \mathsf dt = \frac1\lambda. $$


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