SW VandeCarr said:
It's a very reasonable assumption that the OP was thinking in terms of the standard distributions that are used in statistical applications. Integrating the PDFs of these distributions from negative to positive infinity must yield 1 in order to satisfy the axioms of probability theory. You know this. The kinds of distributions you refer to would not appear to have any applications in the kinds of statistical/probability problems that are generally discussed in this forum.
Given we are referring to these kinds of standard distributions, the probability density is a useful tool when dealing with PDFs. Even when dealing with discrete distributions with large n, using the continuous approximation is often considered preferable.
Finding the mode of a unimodal standard PDF is a straightforward analytic procedure. You know this too. I don't think the line you're pursuing is helpful to the OP. If you persist, I will refer this thread to the moderators.
No, you are wrong and haruspex is correct. The attachment in the OP explicitly states that ##p(x)## is a probability density. It is NOT a probability.
So, this should clear up the miscommunication, unless you also happen to mistakenly believe that a probability density cannot have a maximum value greater than one. I cannot tell from your post if you believe that or if there is just a misunderstanding of what ##p(x)## meant, but in case it is the former: A valid probability density function can absolutely have a maximum value greater than 1. The total area under the curve must be 1, but that does not at all imply the function itself cannot be greater than 1.
Consider a uniform distribution of random variables that can take value only within a very tiny range, say 0.9 to 1.1. The probability density function is p(x) = 5. The area under this curve is 1:
$$\int_{0.9}^{1.1} dx~p(x) = 5 \int_{0.9}^{1.1} dx = 5 (1.1-0.9) = 5 (0.2) = 1.$$
This is a perfectly valid, normalized probability density function. Because the width is so small, the height has to be large in order for the area to come out to 1.
Similarly, take any Gaussian and give it a very small standard deviation. You will find the probability density function exceeds 1 when x is equal to the mean, but the area under the curve will never exceed 1. Again, this is because the peak height has to compensate for the fact that there is very little area under the tails of the Gaussian curve when the standard deviation is very small.
To the OP:
I would interpret the question as expecting you to give the answer "x=0" as the "most probable value" or the "most likely value". Although strictly speaking the probability of getting the value is 0 for any given point, if you draw from this distribution many, many times and generate a histogram which approximates the (unnormalized) distribution, then you will see that you drew many more points in your "x~0" bin than any other points. If you drew from the distribution infinitely many times so that you could make your bins infinitely thin, the greatest fraction of the points that you drew will correspond to "x=0". So, in this sense, the value "x=0" is the most probable value.
Another way to see it is to realize that if you draw from the probability density ##p(x)## infinitely many times, ##f(x_1,x_2) = (Ae^{-\lambda x_1})/(Ae^{-\lambda x_2}) = e^{-\lambda(x_1-x_2)}## is the fraction of times you drew ##x_1## compared to ##x_2##. For ##x_1 = 0 \neq x_2##, you can see that this fraction is always greater than 1, so there are always more draws of "x=0" than any other point. (You don't have to make this argument for every problem like this, it's just to illustrate the sense in which the value of x that maximizes p(x) is the "most probable value").
