# [R] Generate random Bernoulli draws

Berry, Charles ccberry @end|ng |rom uc@d@edu
Fri Jul 6 19:18:58 CEST 2018

```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.

HTH,

Chuck

>>>
>>> Can some one leave me some code lines for implementing this?
>>>
>>>
>>> David
>>>
>>>        [[alternative HTML version deleted]]
>>>
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>>
>>
>
> 	[[alternative HTML version deleted]]
>

```