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I understand that when using dropout, a single neuron can be described using Bernoulli random variable and for a set of neurons it can be described as Binomial random variable

When using Dropout, we define a fixed Dropout probability p for a chosen layer and we expect that a proportional number of neurons are dropped from it.

For example, if the layer we apply Dropout to has n=1024 neurons and p=0.5, we expect that 512 get dropped. Let’s verify this statement:

enter image description here

Thus, the probability of dropping out exactly np=512 neurons is of only 0.025!

why dont we see the probability of 0.5 what am I missing here?

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As you add more Bernoulli trials (neurons in this case) the probability of having a proportion within some small distance of 0.5 (say between 0.49 and 0.51) goes up, but the probability of exactly $n/2$ goes down.

The variance of $X\sim\text{Bin}(n,p)$ is $np(1-p)$. With $p=0.5$ that's $n/4$ ($\sigma_X=\sqrt{n}/2$). So as you add more neurons, the typical (RMS) distance of X from n/2 will go up and the probability of an individual outcome nearest the mean goes down. But the variance of a binomial proportion $\hat{p}=X/n$ is $p(1-p)/n$ ($\sigma_\hat{p}=0.5/\sqrt{n}$), so as you add more trials (neurons) the probability of being within a small interval of the mean proportion goes up.

This is a similar kind of error to the Gambler's fallacy; people expect that an excess of heads will be compensated by an excess of tails in the short run, so that the number of heads will be pushed toward $\frac12 n$. But there's no such effect. In $n$ tosses of a fair coin, let's call the number of heads $N_H$, (similarly for $N_T$, so $N_H+N_T=n$), and let the observed proportion of heads be $p_H$, where $p_H=N_H/n$.

What happens to $E|N_H-N_T|$ as $n$ grows? It doesn't go to zero (as the fallacious gambler reasons) -- as $n\to\infty$, it actually goes to infinity (it grows as $n^\frac12$). However, $E|p_H-p_T|$ does go to zero (as indeed it must, because of the weak law of large numbers).

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  • $\begingroup$ When did we all suddenly start talking taking about deep learning..? $\endgroup$ – Tim Mar 3 '19 at 6:51
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    $\begingroup$ When a lot of the questions started being stats questions? $\endgroup$ – Glen_b -Reinstate Monica Mar 3 '19 at 6:54
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When $n$ gets large, individual probabilities gets smaller, because options are many. For example, if we are dealing with lots of neurons, let's say $n=10^{30}$, $P(Y=5.10^{29})$ will be much much smaller, right? Because, when you have an experiment of such nature, probability is also spreading across the neighbourhood of $np$. In your case, the probability of $Y=511,513 \ ...$ have also some volume. So, you'll consider ranges, not individual probabilities. When you select some from $n$ elements, it won't be that hard to get a number around $np \pm \sqrt{np(1-p)}$, i.e. one deviation away from the mean. Specifically, for one $\sigma=16$, you have a high probability: $P(496\leq Y\leq 528)\approx 0.7$. If we consider $2\sigma$, $P(480\leq Y \leq 544)\approx0.96$. So, it's highly likely that you end up with a number in $[480,544]$. In any case, don't confuse the expected value with individual scores.

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