[R] Generate random Bernoulli draws

Berry, Charles ccberry @end|ng |rom uc@d@edu
Sat Jul 7 01:30:22 CEST 2018

Sorry about the last incomplete post. Accidentally hit send.

Meant to say that I was hoping that a correct, but  obscure response from me would motivate David to step back and think about his problem long enough to see that it has an easy solution.

Sorry if that was out-of-line.


> On Jul 6, 2018, at 4:27 PM, Charles Berry <ccberry using ucsd.edu> wrote:
>> On Jul 6, 2018, at 3:31 PM, Duncan Murdoch <murdoch.duncan using gmail.com> wrote:
>> On 06/07/2018 1:18 PM, Berry, Charles wrote:
>>> A liitle math goes along way. See below.
>>>> On Jul 5, 2018, at 10:35 PM, Marino David <davidmarino838 using gmail.com> wrote:
>>>> Dear Bert,
>>>> I know it is a simple question. But for me, at current, I fail to implement
>>>> it. So, I ask for help here.
>>>> It is not homework.
>>>> Best,
>>>> David
>>>> 2018-07-06 13:32 GMT+08:00 Bert Gunter <bgunter.4567 using gmail.com>:
>>>>> Is this homework?
>>>>> (There is an informal no-homework policy on this list).
>>>>> Cheers,
>>>>> Bert
>>>>> Bert Gunter
>>>>> "The trouble with having an open mind is that people keep coming along and
>>>>> sticking things into it."
>>>>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>>>> On Thu, Jul 5, 2018 at 10:23 PM, Marino David <davidmarino838 using gmail.com>
>>>>> wrote:
>>>>>> Dear All,
>>>>>> I would like to generate N random Bernoulli draws given a probability
>>>>>> function F(x)=1-exp(-2.5*x) in which x follows  uniform distribution, say
>>>>>> x~U(0,2).
>>> If each Bernoulli draw is based on its own draw of x, then
>>> 	rbinom( N, 1, 0.8013476 )
>>> is what you want.
>>> It is left as an exercise for the reader to verify that the constant 0.8013476 is correct up to approximation error, and to prove that such a Bernoulli mixture is also Bernoulli. Perhaps,
>>> 	?integrate
>>> will help.
>>> But if the x's are shared you need to use runif, expm1, and (possibly) rep to produce a vector to be used in place of the prob argument.

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