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I have a question about finding the standard deviation of a discrete version of a Normal Distribution (maybe a binomial distribution?), knowing the mean and the probability density function value of the mean.

For example, using an age recognition algorithm, we get the age of person estimated to be 26 with 40% confidence. Let's assume that the age is distributed like a discrete RV that follows a Normal Distribution, so that its mean is 26 and $\mathbb{P}(26)=0.4$. The age recognition is bounded between 0 and 75 years old, so that $\mathbb{P}(x<0)=0$ and $\mathbb{P}(x>75)=0$.

It seems to me that with this assumption I should be able to compute the probability of other values between 0 and 75. How would I proceed in doing that?

Question edited for clarification:

Let's assume that we have a discretised version of a Normal Distribution: a pmf obtained by taking the integer values of a Normal Distribution within some range (let's say from 0 to 75) and normalizing so that the sum is equal to one. If we know that, from this distribution, P(26)=0.4, how can we calculate the other P(x)? I would like an analytical solution, as a numerical one is easy to obtain.

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    $\begingroup$ What could it mean to be "a discrete RV that follows a Normal Distribution"? Are you talking about a normal distribution that has been rounded to the nearest whole number? (That wouldn't be a normal distribution.) $\endgroup$ – gung - Reinstate Monica Mar 25 '17 at 18:52
  • $\begingroup$ I mean a normal distribution that has been truncated from 0 to 75, discretised on the nearest integer and normalized. $\endgroup$ – u-_-u Mar 25 '17 at 19:13
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I think you're confusing together two separate things.

The first is the distribution of ages (in complete years, say). This would be discrete with values between 0 and some large number (generally there's no specific fixed upper limit on age but in practice you might effectively be able to name an upper bound). If $A$ is the age of a randomly selected person, you can talk about $P(A=a)$ for $a=0,1,....$.

To go with that, there's the estimated ages. These would also (presumably) be discrete with values on a similar domain to the actual ages. Let's call the estimated age of that randomly selected person (and we'll use $H$ for the hestimated age). So we can talk about $P(H=h)$ for $h=0,1,...$.

These ages (actual and estimated) probably won't be even close to a discretized normal.

But the second, different thing is this part: "we get the age of person estimated to be 26 with 40% confidence. Let's assume that the age is distributed like a discrete RV that follows a Normal Distribution, so that its mean is 26 and P(26)=0.4" ... isn't talking about either the distribution of $A$ or the distribution of $H$.

It's talking about the distribution of actual age given the estimated age is 26 (So then we can talk about $p(A=a|H=26)$); it's possible that in some circumstances it might be somewhat like a discretized normal. It's this thing for which we have $P(A=26|H=26) = 0.4$. If you assume that $A$ is a discretized normal where the underlying normal $A^*$ has mean $26$ and standard deviation $\sigma$ then you can compute $\sigma$ from this information since $P(25.5<A^*<26.5)$ $=0.4$ or equivalently $P(\frac{(25.5-26)}{\sigma}<\frac{(A^*-26)}{\sigma}<\frac{(26.5-26)}{\sigma}) = 0.4$ directly implies a value for $\sigma$ [this is now in the form of a routine probability question -- "For $X\sim N(\mu,\sigma^2)$, where $\mu=26$ and $P(-25.5<X<26.5) = 0.4$ find $\sigma$"]. We can continue the previous line of derivation to obtain $\sigma$ formally:

$P(\frac{-0.5}{\sigma}<Z<\frac{0.5}{\sigma}) = 0.4$ ... etc

but first let's clarify the situation and dodge the manipulations along the way.

We've now got it into the form of a question about a standard normal. So what's the z-value for a standard normal that has $0.4$ of the area between $-z$ and $z$?

standard normal density with middle 0.4 of the area shaded

Hint: if the middle purple part is 0.4, what's the area of the two white parts together? So what's the area of the leftmost white part? So what's the area to the left of the value marked "$z$"? Look up your normal tables, or (better still) call a function in any stats package or spreadsheet to find that value.

Now you have the value $z$ for which $0.5/\sigma = z$; solve for $\sigma$.

It seems to me that with this assumption I should be able to compute the probability of other values between 0 and 75. How would I proceed in doing that?

Given the preceding, you can then compute $P(A=a|H=26)$ for other values of $a$. (e.g. $P(A=21|H=26) = 1.18\times 10^{-6}$)

However, it doesn't give you a basis for calculation of the unconditional probability $P(A=21)$ without more information (e.g. if you knew the distribution of $H$ and the conditional distribution at each age, then you should be able to do it, but even if you can assume the conditional distribution of $(A|H=h)$ is discretized normal$(h,\sigma^2)$ you don't have that distribution for $H$).

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  • $\begingroup$ I edited the question. Is it more clear now? $\endgroup$ – u-_-u Mar 26 '17 at 14:07
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    $\begingroup$ 1. That's not what the information at the start of the question is asking about, as I already explained in my answer (it clearly indicates that the information is conditional -- "we get the age of person estimated to be 26" is unambiguous.) ... 2. But in any case, if you insist we treat the information as if it were not conditional, the answer is already covered by the answer I have already given when I did the conditional case - if you treat that as unconditional, then result is the answer I gave for the conditional case -- for example P(A=21) = 1.18 x 10^(-6) (and P(A=31) is the same) $\endgroup$ – Glen_b -Reinstate Monica Mar 26 '17 at 15:15
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    $\begingroup$ If you're just asking me to reopen your question. I didn't close it and I didn't think the question was unclear (in fact I think if anything your edit makes the question less clear), so I can't identify what the reasoning was in closing it -- nobody commented to indicate the problem. I think the question is based on a mistaken premise which is why I answered it the way I did (if I'm wrong then your question as phrased makes little sense). In any case, your edit will place the question in the reopen review queue, so if the people who review it there think it's clear now they'll reopen. $\endgroup$ – Glen_b -Reinstate Monica Mar 26 '17 at 15:23
  • $\begingroup$ Thank you for your answer. Just one clarification needed at this point. You said that the variance can be easily calculated, but I don't see how. Could you clarify that? Or, equivalently, how did you come up with the value 1.18 x 10^(-6) $\endgroup$ – u-_-u Mar 26 '17 at 15:51
  • $\begingroup$ How does this question arise? $\endgroup$ – Glen_b -Reinstate Monica Mar 26 '17 at 17:06

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