Talk:Bernoulli distribution

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MAD error[edit]

Why is the MAD always 1/2 regardless of p? — Preceding unsigned comment added by 206.72.85.89 (talk) 23:31, 18 December 2022 (UTC)[reply]

I think it might be because the average of median is equal to 1/2? 138.246.3.136 (talk) 22:56, 1 May 2023 (UTC)[reply]

I just did the calculations and the Mean Absolute Deviation around the mean is 2p(1-p). However, the Mean Absolute Deviation around the median is 1/2, as the article lists. Very confusing. There doesn't seem to be a consistent standard for what MAD actually means?

See this article for different possible meanings of MAD:

https://en.m.wikipedia.org/wiki/Average_absolute_deviation 73.222.83.173 (talk) 04:17, 6 April 2024 (UTC)[reply]

tacticity question[edit]

does anyone have an idea if Bernoullian probability on the tacticity page is covered in wikipedia and if so where? possibly redirect? thanks in advance V8rik 23:07, 2 August 2005 (UTC)[reply]

Nope. still red links. I have no clue what that is, from context.linas (talk) 17:19, 2 August 2012 (UTC)[reply]

mode error[edit]

i think the mode of the Bernouilli distribution should be zero if p<0.5, should be 1 if p>0.5, and should be equal to "0 and 1" when p=0.5.

Thanks.

Bernoulli kurtosis excess is actually kurtosis?[edit]

The equation given for 'kurtosis excess' is correct.

http://mathworld.wolfram.com/BernoulliDistribution.html gives this formula for the excess but describes it as the kurtosis rather than kurtosis excess as elsewhere.

Although the custom in Wikipedia is to use the term kurtosis for the kurtosis_excess, it is important to avoid confusion.

IMO it would be better to give BOTH in ALL distribution tables to reduce the risk of any more similar muddles.

And to say true kurtosis for kurtosis and excess kurtosis in the text.

BUT I also note that http://mathworld.wolfram.com/Kurtosis.html gives a MUCH simpler formula

 1/(1-p) + 1/p - 6

and this would seem preferable because it is simpler and makes obvious the effect of changing p.

A formula for the 'true' kurtosis is

  1/(1-p) + 1/p - 3

(The kurtosis excess formula given currently is just the binomial with n = 1, and can be simplified).

Paul A Bristow 14:45, 13 December 2006 (UTC) Paul A Bristow[reply]

The article says that the Bernoulli distribution with $p=1/2$ has a lower kurtosis than any other probability distribution. But in fact its kurtosis of $-2$ is the same as that of any distribution that takes two distinct values, each with probability $1/2$ . Fathead99 (talk) 18:23, 5 March 2008 (UTC)[reply]

derivations[edit]

I hope somebody could help me in finding derivations or how to derive the skewness and kurtosis, even link to other sites will be much appreciated. —Preceding unsigned comment added by Student29 (talk • contribs) 19:31, 16 January 2008 (UTC)[reply]

Why no median?[edit]

Is there any particular reason why the Bernoulli distribution lacks a defined median? I would think that the median is 1 if p > 0.5, 0 if p < 0.5, 0.5 if p = 0.5. 169.232.78.24 (talk) —Preceding comment was added at 21:24, 7 February 2008 (UTC)[reply]

I think the median is 1 if p > 0.5 and 0 if p < 0.5, but it does not exist of p = 0.5. -- NaBUru38 (talk) 16:23, 11 February 2008 (UTC)[reply]

CDF[edit]

I'm from Spanish Wikipedia so my english is not good. Sorry about that. I think there's a mistake in the CDF. Here it's defined the CDF for k real, but k only can be 0 or 1. I think that CDF should be (1 - p) for k = 0 and 1 for k = 1. --201.212.140.201 (talk) 18:47, 13 April 2009 (UTC)[reply]

I agree with the previous comment. Needs fixing. —Preceding unsigned comment added by 130.91.169.182 (talk) 22:44, 15 April 2010 (UTC)[reply]

Why no Simple Language?[edit]

Attach more links to concepts explained in a way that people who forgot calc 1 a decade ago can understand. Also, explain how integrals and other complex mathematical terms apply to this concept. Please feel free to edit this into a more editor-friendly lang.

Mistake in formula?[edit]

The article says the Bernoulli distribution can be expressed as

f(k;p)=p^{k}(1-p)^{1-k}\!\quad {\text{for }}p\in \{0,1\}

.

Surely this should read

f(k;p)=p^{k}(1-p)^{1-k}\!\quad {\text{for }}k\in \{0,1\}

.

instead.

129.173.212.177 (talk) 17:18, 20 May 2010 (UTC)[reply]

Also

k

should be defined. — Preceding unsigned comment added by 159.54.131.7 (talk) 14:35, 10 August 2011 (UTC)[reply]

"Number of records"[edit]

Just added this section, not sure if the wording is correct — Preceding unsigned comment added by Herix (talk • contribs) 14:16, 7 October 2011 (UTC)[reply]

Bournoulli vs. Binomial[edit]

My understanding of the difference between a Bernoulli and a binomial trial, model, distribution, etc. is that in a binomial model, all the subjects are assumed to have the same outcome probability (p is the same for all k subjects), while in the more general Bernoulli model, p(i) may vary for each subject. Thus, a binomial model (etc.) is a kind of Bernoulli model, but not vice-versa unless k=1. I notice the Wikipedia articles appear to treat the two as identical. --69.115.221.42 (talk) 12:53, 12 June 2012 (UTC)[reply]

You'd have to provide a reference for that, because I don't think it makes any sense. Every book I've seen makes it clear that Bernoulli-anythings are shift-invariant, stationary processes. The only way that they can be stationary is if p(i) is exactly the same for all shifts i; that is the definition of "stationary". linas (talk) 17:16, 2 August 2012 (UTC)[reply]

Confused probability distributions and random variables[edit]

The very first sentence seems to confuse probability distributions and random variables:

In probability theory and statistics, the Bernoulli distribution, named after Swiss scientist Jacob Bernoulli, is a discrete probability distribution, which takes value 1 with success probability $p$ and value 0 with failure probability $q=1-p$ .

Obviously the distribution can only be identified with a function that takes the values $p$ and $1-p$ for 1 and 0, respectively. Perhaps it would be best to just insert "... is the probability distribution of a random variable that takes ..." Seattle Jörg (talk) 13:41, 11 February 2014 (UTC)[reply]

E(xi)=p then for n tries E(s)=n*p — Preceding unsigned comment added by Ahyorya (talk • contribs) 16:17, 29 August 2014 (UTC)[reply]

why is the chart in german?[edit]

also, does this really need a chart? it' s just a generalized coin flip, it feels unnecessary. --3thanguy7 (talk) 19:58, 6 February 2022 (UTC)[reply]

The chart itself is confusing because it plots 3 separate distributions onto one graph and also separates the lines/bars so they appear offset. It would be much clearer just to show one distribution.

The point of the chart is to show the distribution as point masses of a PMF. You can actually treat discrete and continuous distributions together with a appropriate measure integral. Wqwt (talk) 23:18, 14 February 2022 (UTC)[reply]