$$\sum_{i=1}^n bernoulli(p) = binomial(n,p) \approx \mathcal N(np, np(1-p)) = \sum_{i=1}^n \mathcal N(p, p(1-p))$$

Can I conclude that $\mathcal N(p, p(1-p))$ could represent an approximation of $bernoulli(p)$?

In particular, given $n$ binary RVs $X_i$ then a possible naive factorization of $P(X_1, X_2, \ldots, X_n)$ is $P(X_1) P(X_2) \ldots P(X_n)$.

Since all the RVs are binary, then they can be modeled as Bernoulli RVs.

In case I am not interested in the exact probability of the joint, the can I use the normal distribution to approximate each bernoulli variable?

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    $\begingroup$ Just how poor an approximation can you tolerate? After all, you are proposing to use a continuous distribution to approximate a discrete distribution having two values: except perhaps for some very special calculations, such an approximation would seem to have no advantages. $\endgroup$ – whuber Dec 5 '18 at 19:52
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    $\begingroup$ When you approximate a binomial with a normal you are using large sample (large n) property. Although there are no hard rules on how large n should be for the sample to be large, I think it is safe to say that a sample of 1 is not large. $\endgroup$ – Jesper for President Dec 5 '18 at 19:52

Let's analyze the error.

The figure shows plots of the distribution function of various Bernoulli$(p)$ variables in blue and the corresponding Normal distributions in Red. The shaded regions show where the functions differ appreciably.

Figure 1

(Why plot distribution functions instead of density functions? Because a Bernoulli variable has no density function. The densities of good continuous approximations to Bernoulli distributions have huge spikes in neighborhoods of $0$ and $1.$)

No matter what $p$ may be, for some values of $x$ the difference between the two distribution functions will be large. After all, the Bernoullli distribution function has two leaps in it: it jumps by $1-p$ at $x=0$ and again by $p$ at $x=1.$ The Normal distribution function is going to split the greater of those two leaps into two parts, whence the larger of the two vertical differences--the largest error--must be at least $1/4.$ In fact, it's always greater even than that.

Here is a plot of the maximum difference between the two functions, as it depends on $p:$

Figure 2

It is never smaller than $0.341345,$ attained when $p=1/2.$ Because probabilities all lie between $0$ and $1,$ this is a substantial error. It is difficult to conceive of circumstances where this approximation would be acceptable, except perhaps when $x\lt 0$ or $x\gt 1:$ but then why use a Normal distribution at all? Just approximate those values as $0$ and $1,$ respectively, without any error at all.

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I don't think you can conclude that N(p,p(1−p)) could represent an approximation of bernoulli(p). First of all, for a bernoulli variable, a random sample could only be 0 or 1, on the other hand, the range of normal variable could be from -inf to inf. Secondly, If we have a random distribution with mean p, and variance p(1-p), once we draw lots of samples from this distribution and add them together, their summation distribution will also follow a normal distribution with mean np and variance np(1-p) due to central limit theorem. For sure we can't say the random distribution represents an approximation of bernoulli(p)...

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    $\begingroup$ The "Secondly" part is unclear, this is what OP is saying, isn't it? Why should this be argument against it? $\endgroup$ – Tim Dec 5 '18 at 20:56
  • $\begingroup$ Hi Tim, I don't explain it well, I was trying to claim that if the argument is correct, then we could use a random distribution to approximate the bernoulli distribution, which is clearly not correct. let me give you a example, a normal distribution with mean 1 and variance 1, also a poisson distribution with mean 1 and variance 1. the summation of N normal samples is another normal distribution with mean N and variance N, the summation of N poisson samples also follows normal distribution with mean N and variance N. (N is large) but apparently poisson (1) can't approximate normal(1, 1) $\endgroup$ – Yang Song Dec 6 '18 at 19:41

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