Timeline for Expected values of two random variables related to a simple urn problem
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| S Feb 12, 2019 at 20:00 | history | bounty ended | CommunityBot | ||
| S Feb 12, 2019 at 20:00 | history | notice removed | CommunityBot | ||
| Feb 11, 2019 at 8:13 | answer | added | JMP | timeline score: 0 | |
| Feb 5, 2019 at 16:04 | answer | added | Carlo Beenakker | timeline score: 2 | |
| Feb 5, 2019 at 15:16 | comment | added | Andrea Prunotto | @CarloBeenakker Yes, maybe this is a better approach. How can I proceed, then? Can I use a similar approach for the other event? Thanks for your help! | |
| Feb 5, 2019 at 9:10 | comment | added | Carlo Beenakker | I'm afraid the question as phrased has no answer; a different question does: "Repeatedly draw a ball with replacement, stop when you have $n$ black balls. What is the expected number of balls drawn?" --- is that a question that would interest you? | |
| Feb 5, 2019 at 5:59 | comment | added | Andrea Prunotto | @CarloBeenakker True. The obscure point (for me) is that $X=n$ does not occur with probability $1$. I thought that, since $X=k$ occurs with $P(X=k)=0$ for each $k\neq n$, and $X=n$ with $P(X=n)=\frac{b^n}{c^n}$, then we could use the distribution of $P(X)$ to define the expected value. I understand that something is wrong here, but I don't see exactly where. | |
| Feb 4, 2019 at 19:51 | comment | added | Carlo Beenakker | if, as you say, the variable $X$ can assume only the value $𝑛$, then how can its expected value be any different from $n$? | |
| S Feb 4, 2019 at 18:24 | history | bounty started | Andrea Prunotto | ||
| S Feb 4, 2019 at 18:24 | history | notice added | Andrea Prunotto | Authoritative reference needed | |
| Jan 30, 2019 at 22:45 | comment | added | Andrea Prunotto | Sorry, I meant $P(Y=k)=1-\frac{(u-b)^k}{u^k}$. | |
| Jan 30, 2019 at 22:38 | comment | added | Andrea Prunotto | @IosifPinelis A similar approach (sorry if it is not well stated) applies for the other event. Among the $n$ trials, each of them can be the one in which the event "to get at least one black ball" occurs. Which trial? Randomly, any among $Y=1,2,\ldots n$. What is the probability to get at least one black ball at the $k$-th trial? Well, it is $1-\frac{(u-b)^n}{u^n}$. I would say that this is $P(Y=k)$. | |
| Jan 30, 2019 at 22:33 | comment | added | Andrea Prunotto | @IosifPinelis Among the $n$ trials, one and only one of them can be the one in which the event $E$ occurs. Well, that trial can be only the last one! Therefore, if we denote with $X$ the number of trials needed to get $n$ black balls, well $X$ can be only equal to $n$. What is the probability that this $n$-th trials is the one related to the extraction of exactly $n$ black balls, out of $u$? Welll, $\frac{b^n}{u^n}$. | |
| Jan 30, 2019 at 22:26 | comment | added | Andrea Prunotto | @IosifPinelis I see. I'm sorry if I couldn't be more clear. Maybe I can do the other way round, i.e. to tell you (in non-mathematical terms) what I am looking for, and then you can help me to formulate better the question. | |
| Jan 30, 2019 at 17:32 | comment | added | Iosif Pinelis | Sorry, I still don't understand your definitions. Perhaps you can state them in purely formal terms (such as independent identically distributed random variables, with certain distributions) and completely eschew such non-mathematical terms as "draw", "attempt", "trials", etc. | |
| Jan 30, 2019 at 16:33 | comment | added | Andrea Prunotto | @IosifPinelis We have $n$ attempts. $X$ is the number of times I should draw a ball from the urn in order to get $n$ black balls. $Y$ is the number of times I should draw a ball from the urn in order to get at least one black ball. In the first case, I always need to do $X=n$ attempts, and the probability that all the balls are black is $P(X=n)=\frac{b^n}{u^n}$. In the second case, I can do $Y=\{1,2,3\ldots n\}$ attempts, because in all these cases I have a probability to get at least one black ball, which is $P(Y=k)=1-\frac{(u-b)^k}{u^k}$. Is this ill posed? | |
| Jan 30, 2019 at 14:34 | comment | added | Iosif Pinelis | The definitions of $X$ and $Y$ are still unclear. In particular, what will be the value of $X$ if none of the first $n$ balls is black? More generally, can you define $Y$ and $Y$ formally, without using terms such as "trials", "bets", "winning", etc.? | |
| Jan 30, 2019 at 5:30 | comment | added | Andrea Prunotto | @IosifPinelis Suppose that I bet to get at least one black ball in $n=10$ trials (event $L$). I extract $1,2\ldots k$ balls from the urn. If at the trial $k=3$ I get a black ball, I have already won the bet, no matter the following trials. This means that the trial $k=3$ was successful for the event $L$, and $Y=3$ | |
| Jan 30, 2019 at 2:42 | comment | added | Iosif Pinelis | In your definitions of of $X$ and $Y$, what do you mean by "trials" and "a success for the event"? | |
| Jan 29, 2019 at 19:36 | history | edited | Andrea Prunotto | CC BY-SA 4.0 |
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| Jan 29, 2019 at 18:52 | comment | added | Andrea Prunotto | @DieterKadelka Thanks for your comment. | |
| Jan 29, 2019 at 18:26 | comment | added | Dieter Kadelka | If I understand your question (the desription does not make much sense for me), the random variables $X$ and $Y$ are geometrically distributed. These random variables are not bounded by $n$, so your calculations cannot be correct. Please google for geometric distribution. | |
| Jan 29, 2019 at 18:24 | history | edited | Andrea Prunotto | CC BY-SA 4.0 |
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| Jan 29, 2019 at 12:07 | history | edited | Andrea Prunotto | CC BY-SA 4.0 |
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| Jan 29, 2019 at 10:58 | history | asked | Andrea Prunotto | CC BY-SA 4.0 |