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In discrete distribution when we plot PMF the Y axis is probability. In continuous distribution when we plot PDF the Y axis is density (probability is the area under the curve). So, we learn that density values are not probability values.

But what happens when I approximate binomial with normal distribution. Consider an example case of B(100, 0.5). So, Mu=50, sigma=5. I calculated both binomial and normal distributions with these parameters. Below is the plot. For binomial distribution my Y axis is probability but for normal distribution it is density. But numerically the values practically overlap. Obviously, probability in a point for normal distribution is still 0. Does this makes sense to you?

Thanks!

enter image description here

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  • $\begingroup$ I think that depends on the normalization for the histogram. If you choose a bin width different from 1, you won’t obtain a good fit with the Gaussian unless you normalize the histogram ( usually an option in CAS). In this latter case the probability is the area under the histogram $\endgroup$ – lcv Dec 22 '18 at 12:36
  • $\begingroup$ In order to connect with the valid answers that you got, note that for general bins $P(X=k) \approx P(k-\Delta k/2 < Y < k+ \Delta k/2) \Delta k$, where $\Delta k$ is the size of the bin. $\endgroup$ – lcv Dec 22 '18 at 12:41
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    $\begingroup$ I see. Thanks a lot for the help. $\endgroup$ – John Dec 22 '18 at 14:34
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What you observe is that: $$P(X=k)\approx P\left(k-\frac12<Y<k+\frac12\right)=\int_{k-\frac12}^{k+\frac12}f_Y(y)dy$$where $X$ is binomial and $Y$ is normal.

So actually expressions like $\int_{k-\frac12}^{k+\frac12}f_Y(y)dy$ where $f_Y(y)$ is a PDF can be recognized here as probabilities.

In a suitable situation some $x_0\in(k-\frac12,k+\frac12)$ may exist that satisfies $$P(X=k)=\int_{k-\frac12}^{k+\frac12}f_Y(y)dy=\int_{k-\frac12}^{k+\frac12}f_Y(x_0)dy=f(x_0)$$

This however does not prevent that a PDF can take values in $(1,\infty)$, and this is not possible for a probability.


edit:

Let me say also that lots of PMF's (not all) actually induce a PDF.

For instance let $p$ denote the PMF of a random variable $X$ that takes values in $\mathbb Z$.

The function $f$ prescribed by $f(x)=\lfloor x\rfloor$ is then a PDF.

This because it is measurable, non-negative and satisfies: $$\int f(x)dx=\sum_{n\in\mathbb Z}p(n)=1$$ It is the PDF of $X+U$ where $X$ and $U$ are independent and $U$ has uniform distribution on $[0,1)$.

By picturing these PMF and PDF will have the same looks except that the PMF only takes positive values on $\mathbb Z$ and not on $\mathbb R-\mathbb Z$.

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  • $\begingroup$ Thanks, but I am not sure that I fully understand your answer. In my specific case the values on Y axis are the same for both discrete and continuous case. How can the PDF take (1, infinity) ? $\endgroup$ – John Dec 22 '18 at 11:38
  • $\begingroup$ When I say that PDF's can take values in $(1,\infty)$ then I am not talking specifically about the PDF in your quesion, but about PDF's in general. $\endgroup$ – drhab Dec 22 '18 at 11:41
  • $\begingroup$ I see. Thank you for the help. $\endgroup$ – John Dec 22 '18 at 14:35
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    $\begingroup$ @John Are you aware that you can accept answers? Accepting an answer is a kind of appreciation for the people who take time to help you. Accept the answer which fits at best to your question. $\endgroup$ – callculus Dec 22 '18 at 18:03
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Note that: $$P(X=k)\approx P\left(k-\frac12<Y<k+\frac12\right)=\\ =P\left(\frac{\left(k-\frac12\right)-50}{5}<Z<\frac{\left(k+\frac12\right)-50}{5}\right)=\\ P\left(\frac k5-10.1<Z<\frac k5-9.9\right),$$ where $X$ is the discrete r.v., $Y$ is the continuous normal r.v. and $Z$ is the continuous standard normal r.v.

Obviously, probability in a point for normal distribution is still $0$.

The thing is the probability of the binomial r.v. $X=k$ approximately corresponds to the density (or area) under the normal curve for the interval $(k-\frac12, k+\frac12)$. As the width of the rectangle gets smaller and the height of the rectangle gets bigger, the area does not change and is approximately equal to the probability of the normal (standard normal) random variable in the specified interval.

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  • $\begingroup$ This is only valid if the size of the bins is 1. I think this is where the OP’s confusion originates. See my comment. $\endgroup$ – lcv Dec 22 '18 at 12:44
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    $\begingroup$ Thanks you both for the help. $\endgroup$ – John Dec 22 '18 at 14:34

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