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This is my probability density function (pdf) $$pdf = e^{-\frac{r}{\lambda}} \frac{1}{\lambda}$$

I want to find the expected value (EV) from $0<r \leq r_0$, $r$ is the random variable.

My first attempt is $EV_1 - EV_2$ (see below calculation), but this does not work.

The solution from the paper is $EV_3 + EV_4 = \lambda - \lambda e^{-\frac{r_0}{\lambda}}$

my questions:

  1. Are integration limits used to find expected value from a pdf inclusive (or exclusive)? it seems like the lower limit is inclusive but the upper limit is exclusive. Any reference that I can look about this?
  2. What is the interpretation of $EV_4$?

Here are some values that I calculated. $$EV_1 = \int_0^\infty r \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = \lambda $$

$$EV_2 = \int_{r_0}^\infty r \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = r_0 e^{-\frac{r_0}{\lambda}} + \lambda e^{-\frac{r_0}{\lambda}} $$

$$EV_3 = \int_{0}^{r_0} r \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = \lambda - r_0 e^{-\frac{r_0}{\lambda}} - \lambda e^{-\frac{r_0}{\lambda}} $$

$$EV_4 = \int_{r_0}^\infty r_0 \left(e^{-\frac{r}{\lambda}} \frac{1}{\lambda} \right) dr = r_0 e^{-\frac{r_0}{\lambda}} $$

This question is actually a physical formulation that I am reading. SO, if you need more explanation about the problem, let me know.

EDIT: These are the 3 references I used. They are discussing the same thing in the screenshots I attached.

  1. https://doi.org/10.1080/00018735200101151 https://doi.org/10.1080/00018735200101151

  2. https://doi.org/10.1098/rspa.1950.0077 https://doi.org/10.1098/rspa.1950.0077

  3. https://doi.org/10.1098/rspa.1950.0107 https://doi.org/10.1098/rspa.1950.0107

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  • $\begingroup$ Your pdf doesn't integrate to 1 over 0 to $r_0$. You need to fix that first. $\endgroup$ – JimB Mar 16 at 18:17
  • $\begingroup$ why should my pdf integrate to 1 over 0 to $r_0$? it does integrate to 1 over 0 to $\infty$. $\endgroup$ – Codelearner777 Mar 16 at 18:22
  • $\begingroup$ "I want to find the expected value (EV) from 0<r≤r0, r is the random variable." $\endgroup$ – JimB Mar 16 at 18:37
  • $\begingroup$ I do not want to find the EV of the whole possible r.v., but limited to $0<r\leq r_0$. Maybe the word mean or average is more proper? $\endgroup$ – Codelearner777 Mar 16 at 18:42
  • $\begingroup$ You seem to be using very non-standard terminology. (Is this terminology from some electrical engineering statistics course?) An expected value requires a pdf to integrate to 1 (or sum to 1 if discrete). Otherwise you're just finding disjoint parts of an integral (that is an expected value). Finding the expected value of $r$ given that $0< r\leq_0$ is labeled $E(r|0<r\leq r_0)$. $\endgroup$ – JimB Mar 16 at 19:11
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From the references you give, it is now clear that you want the mean of a censored distribution rather than a truncated distribution. Specifically, if you have a random variable $r$ with pdf $\frac{\exp \left(-\frac{r}{\lambda }\right)}{\lambda }$, then you want the mean of the censored variable $s$ which is $s=r$ if $0<s\leq r_0$ and $s=r_0$ if $r>r_0$.

$$E(s)=\int_0^{r_0} \frac{r \exp \left(-\frac{r}{\lambda }\right)}{\lambda } \, dr + \int_{r_0}^{\infty } \frac{r_0 \exp \left(-\frac{r}{\lambda }\right)}{\lambda } \, dr$$

$$=\left(\lambda -e^{-\frac{r_0}{\lambda }} (\lambda +r_0)\right)+\left(r_0 e^{-\frac{r_0}{\lambda }}\right)=\lambda -\lambda e^{-\frac{r_0}{\lambda }}$$

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  • $\begingroup$ TIL new things. thanks. $\endgroup$ – Codelearner777 Mar 16 at 22:00
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    $\begingroup$ Me, too. I've never hear of TIL before (until now and googling it). I should have assumed a large age difference. (I'm retired.) And if you really need an interpretation of the two parts that might be the following: (1) the conditional mean $E(r|0<r\leq r_0)$ times the probability of $0<r\leq r_0$ and (2) $r_0$ times the probability that $r>r_0$. $\endgroup$ – JimB Mar 16 at 22:25

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