For a uni-variate normal distribution with mean "-3", I draw a probability density function curve for this, with y-axis = probability density. Can this curve pass through the co-ordinate (x,y) = (3, 0.2) for any value of standard deviation(0 to infinity)?The solution should be such that almost anyone can do in their head, without using pen and paper, computer etc.

I assume I can perform simple estimated calculations.Gaussian distribution is 1/sqrt(2*pi)/sigma exp[-(x-mu)^2/(2*sigma^2)]. Its maximum is 1/sqrt(2*pi)/sigma and half-maximum is the half of it... E.g. for std equal to 1, 1/sqrt(2*pi)*0.5 =~ 0.2, which is around the y coordinate of your point. You can also estimate the half width of the distribution at half maximum: for mu = 0, 1/2 = exp(-1/2 w^2/sigma^2), ln(0.5) = -log(2) = -1/2 w^2/sigma^2, w = sigma*sqrt(2*ln(2)). You can see that the maximum of the Gaussian scales with sigma like 1/sigma, while its half width grows linearly. For sigma = 1, the half width is slightly more than 1. If you stretch the distribution by increasing sigma to 2, the half width will be twice bigger, but the maximum of the distribution will already drop to 0.2. This shows that there's no chance to get to x=3 at y=0.2 for a Gaussian with maximum at x=mu=-3 (which should have been intuitively obvious since the beginning).The simpler the problem, the more words it takes to describe it

I am not sure if I understood your last part, the intuitive conclusion. For example when sigma = 0.2, ymax =~ 2, I wasn't able to use same logic to prove that it doesn't pass through (3, 0.2)Btw, the solution I had in mind was extremely simple, which requires no log(), sqrt() or exp().

Did you mean y-axis = probability density?

Thanks ymous, yes y-axis = prob densityI corrected the error, ymax =~2, when sigma = 0.2

If the PDF of a N(-3,sigma) is 0.2 at x=+3, then the PDF at x=-9 is also 0.2 and the PDF at x = -3 must be greater than or equal to 0.2. But this implies that the integral on -9 ≤ x ≤ 3 will exceed 2.4 which is impossible.By this simple logic that the integral of the PDF over -9 ≤ x ≤ 3 must be less than one implies that the PDF at x=+3 must be less than 1/12. Further logic concerning the total probability mass in the tails and logic concerning the relative PDF in the mean versus the outlining point (UV's argument) would provide more restrictive upper bounds.

@Traden4Alpha, that was the answer I was looking for@outrun, I am not sure if I got your pointy-axis was probability density, and not probability

You know that atQuote I draw a probability density function curveI thought you meant you wanted to draw a random pdf, and I was starting to think this'd be sick.

QuoteOriginally posted by: Traden4AlphaIf the PDF of a N(-3,sigma) is 0.2 at x=+3, then the PDF at x=-9 is also 0.2 and the PDF at x = -3 must be greater than or equal to 0.2. But this implies that the integral on -9 ≤ x ≤ 3 will exceed 2.4 which is impossible.Yeah, but you used multiplication... I'm kidding Cute solution.

QuoteOriginally posted by: UltravioletQuoteOriginally posted by: Traden4AlphaIf the PDF of a N(-3,sigma) is 0.2 at x=+3, then the PDF at x=-9 is also 0.2 and the PDF at x = -3 must be greater than or equal to 0.2. But this implies that the integral on -9 ≤ x ≤ 3 will exceed 2.4 which is impossible.Yeah, but you used multiplication... I'm kidding Cute solution.Did I? 0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2

QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: UltravioletQuoteOriginally posted by: Traden4AlphaIf the PDF of a N(-3,sigma) is 0.2 at x=+3, then the PDF at x=-9 is also 0.2 and the PDF at x = -3 must be greater than or equal to 0.2. But this implies that the integral on -9 ≤ x ≤ 3 will exceed 2.4 which is impossible.Yeah, but you used multiplication... I'm kidding Cute solution.)Did I? 0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2 This too has an implicit multiplication, 0.2*1 + 0.2*1 + ...

QuoteOriginally posted by: FritzJacobQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: UltravioletQuoteOriginally posted by: Traden4AlphaIf the PDF of a N(-3,sigma) is 0.2 at x=+3, then the PDF at x=-9 is also 0.2 and the PDF at x = -3 must be greater than or equal to 0.2. But this implies that the integral on -9 ≤ x ≤ 3 will exceed 2.4 which is impossible.Yeah, but you used multiplication... I'm kidding Cute solution.Did I? 0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2+0.2 This too has an implicit multiplication, 0.2*1 + 0.2*1 + ...;-)Or maybe I added two line segments:

if you happen to know that when one draws samples x_i from a normal distribution with mean of mu, sqrt(sum{from i=1 to n}(x_i-mu)^2/n) is the maximum likelihood estimator (though biased) of the standard deviation, then this precisely states that the p.d.f. at this particular single sample x=3 reaches maximum when the standard deviation is 3-(-3)=6. it follows that the exact bound is 1/(sqrt(2*pi*e)*6), thus there is some room (about a factor of 2) beyond the normalization argument above.(note this can be extended to get the maximum of the product of the p.d.f. at several points with unknown mean and standard deviation. it's related to the differential entropy of the normal distribution)the above argument of course uses differentiation, so here is the less-restrictive follow-up question, can you improve T4A's bound of 1/12 without using calculus?