This question was given to me as a review for an upcoming exam.

Find $P(2 \leq X \leq 3)$ if $X$ is a continous random variable with pdf: $$f_X(x) = \begin{cases} xe^{-x} & x\geq 0 \\ 0 & x \lt 0 \end{cases} $$ My work: $$P(2 \leq X \leq 3) = \int_{2}^{3}xe^{-x}dx$$

Integration by parts:


$$\int_{2}^{3}xe^{-x} dx = -xe^{-x}\Big|^3_2-\int_{2}^{3}-e^{-x}dx\\ =\frac{-4}{e^3} + \frac{3}{e^2} \approx 0.2069$$

Did I miss anything or do something incorrectly?

Also, I had a quick question regarding a different case, let's say problem was asking for $P(-1 \leq X \leq 3)$ instead, would we do this instead?: $$P(-1 \leq X \leq 3)= \int_{-1}^{0}0 dx + \int_0^3xe^{-x}dx$$

Obviously the left integral will be 0 in this case, I'm just wondering if we split up the integrals with addition if the range we are looking for is in two different domains of the PDF.

  • 1
    $\begingroup$ To your quick question -- yes $\endgroup$ – Minus One-Twelfth Mar 23 at 1:55
  • $\begingroup$ Both parts are correct. $\endgroup$ – Kavi Rama Murthy Mar 23 at 5:26
  • $\begingroup$ Perhaps see Wikipedia. This is the PDF of a RV with dist'n $\mathsf{Gamma}(shape=2, rate=1).$ In R: code diff(pgamma(c(2,3), 2, 1)) returns 0.2068576. Also, pgamma(3, 2, 1) returns 0.8008517. [pgamma is the CDF of a gamma distribution.] Because the support is the positive half line, you could just find $P(0 \le X \le 3).$ $\endgroup$ – BruceET Mar 23 at 17:07

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