Timeline for Expected Value of Exponential CDF
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 13, 2021 at 19:43 | comment | added | Subrata Pal | @Paul1911, You're most welcome... | |
| Jul 13, 2021 at 19:06 | comment | added | Paul1911 | @SubrataPal This is really helpful, thank you! | |
| Jul 13, 2021 at 18:53 | vote | accept | Paul1911 | ||
| Jul 12, 2021 at 15:34 | comment | added | Subrata Pal | A much simpler way to calculate the expectation without bothering about all these would be: $\int_0^\infty P(X>x)dx=\int_0^\infty (1-F(x))dx=E[X]$ The simpler proof is here, but the proof is exactly not applicable for your problem as it assumes the existence of a density. A rigorous proof is here | |
| Jul 12, 2021 at 15:28 | comment | added | Subrata Pal | And the expectation would be: $0.28/0.5+0.71/0.25+0.01*0=3.4$, which is the same as yours because the 3rd random variable is degenerate at 0. | |
| Jul 12, 2021 at 15:25 | comment | added | Subrata Pal | More technically, the absolutely continuous distributions are dominated by the Lebesgue measure and the discrete distributions are dominated by the counting measures. This can neither be dominated by counting measure, nor by the Lebesgue measure, but a mixture of the two. | |
| Jul 12, 2021 at 15:23 | comment | added | Subrata Pal | Hi @Paul1911, it means that your cdf is a mixture of two exponential distributions with parameters 0.5 and 0.25 respectively, and one degenerate discrete random variable(X=0) with weights 0.28, 0.71, 0.01. This distribution does not have a probability density or a probability mass function in the usual sense as it is a mixture of both continuous and discrete random variables. So simply there is no pdf/pmf - but it has a cdf. | |
| Jul 12, 2021 at 13:24 | comment | added | whuber♦ | In this mixture, one of the weights is negative. Such things arise among sums of Gamma variables, for instance. But discussing that would take us far from the intent of this question, which is a basic exercise in integration only: the answer equals $\int_0^\infty f(x)\mathrm{d}x.$ It's unnecessary (and extra work) even to compute the pdf. | |
| Jul 12, 2021 at 7:59 | comment | added | Paul1911 | Hi Subrata, thank you first of all. Looking at your last paragraph, I am not quite sure of the implications for me. The numbers were not a typo, they indeed do not sum up to 1. But I am not sure, what this now means for me to get a correct PDF/CDF - is it only the factor of 0.01 which must be added? Also, you are correct, I forgot the range. It is $[0,\infty)$. | |
| Jul 12, 2021 at 0:51 | review | First posts | |||
| Jul 12, 2021 at 6:03 | |||||
| Jul 12, 2021 at 0:45 | history | answered | Subrata Pal | CC BY-SA 4.0 |