Timeline for Convergence in probability and distribution
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 17, 2019 at 12:01 | vote | accept | sam_rox | ||
| Oct 25, 2018 at 10:43 | history | edited | kjetil b halvorsen♦ |
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| Jan 14, 2018 at 17:11 | history | edited | kjetil b halvorsen♦ |
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| Apr 22, 2017 at 22:49 | answer | added | Dilip Sarwate | timeline score: 1 | |
| Apr 22, 2017 at 15:32 | review | Close votes | |||
| Apr 22, 2017 at 22:01 | |||||
| Apr 22, 2017 at 15:10 | answer | added | kjetil b halvorsen♦ | timeline score: 1 | |
| Mar 12, 2017 at 3:22 | comment | added | sam_rox | @MichaelHardy Thank you for the explanation. This was from a article and they have obtained $E(K)={1\over\beta}$. For this to happen it should be $\Pr(K=k) = (1-\beta)^{k-1}\beta \text{ for } k = 1,2,3,\ldots;$ and not $P(K=k)=(1-\beta)^k\beta ; k=1,2,3,...$ | |
| Mar 12, 2017 at 2:55 | comment | added | Michael Hardy | If $K$ is the number of independent trials needed to get one success, with probability $\beta$ of success on each trial, then the event $K\ge k$ is the same as the event of failure on the first $k-1$ trials; therefore $\Pr(K\ge k) = (1-\beta)^{k-1}. \qquad$ | |
| Mar 12, 2017 at 2:52 | comment | added | Michael Hardy | You need $\Pr(K=k) = (1-\beta)^{k-1}\beta \text{ for } k = 1,2,3,\ldots;$ with $k$ rather than $k-1$ in that exponent, the probabilities will not add up to $1. \qquad$ | |
| Mar 12, 2017 at 2:51 | comment | added | Michael Hardy | To say that $X$ is exponentially distributed with expected value $1$ is equivalent to saying $\Pr(X>x) = e^{-x}$ for $x\ge0.$ That's a somewhat simpler expression than that for $\Pr(X\le x).\qquad$ | |
| Mar 12, 2017 at 2:36 | history | edited | Michael Hardy | CC BY-SA 3.0 |
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| Mar 11, 2017 at 23:37 | history | asked | sam_rox | CC BY-SA 3.0 |