Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Guys I am having trouble with the standard normal distribution.


We know the X values run from approx $-\infty$ to $+\infty$ but what are the y values?? The normal distribution takes two parameters $\mathcal{N}(\mu, \sigma^2)$ but what is the range of y?

$y>0$ obviously and the "y" will depend on the mean and variance you picked as $y=\frac{\exp(-z^2)}{\sqrt{2\pi\sigma^2}}$. But I have trouble understanding what it means. If I take the S&P500 and I difference the series (SPX-SPX(-1)) the histogram of the returns will have an approximate normal distributions and will list out the number of times I have a return of -1%,-.5%,0%,.5%, 1% , etc throughout the history. So is the "y" of the normal distribution the number of times I have had that x as a value? Should I think of the normal distribution in practical terms the number of times that one point event has occurred? I look at some normal distributions and the Y ranges from 0-4, others I see the y ranging from 0 to 1, as a probability should. I know the area underneath the curve should sum to 1 but shouldnt the y values always be less than 1?


Thanks guys!

share|improve this question

2 Answers 2

You may be thinking of the cumulative distribution function, which takes on all values in the interval $(0,1)$. Or else you may be thinking of the (probability) density function $$\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}},$$ the familiar "bell-shaped" curve. This density function is positive, but not necessarily between $0$ and $1$. It reaches a maximum when $x=\mu$. The maximum value (what your post would call the maximum $y$-value) is $\dfrac{1}{\sqrt{2\pi}\sigma}$. The range of the density function is the interval $\left(0,\frac{1}{\sqrt{2\pi}\sigma}\right]$.

In particular, when $\sigma$ is small, the maximum value can be quite large: the density function reaches a sharp high peak. If $\sigma$ is large, the density function, though still characteristically bell-shaped, is flat and low. The area under the density curve, and above the $x$-axis, is always $1$. So if the density function is near $0$ very soon (small variance,) it is intuitively clear that the curve must reach quite high.

Remark: Let $f(x)$ be our probability density function. Then for small $h$, the probability that our random variable lies between $x$ and $x+h$ is approximately $hf(x)$. In that sense, you can pick up a pretty good picture of $f(x)$ if you have a largish number of data points.

share|improve this answer
I see, so the "Y" is unbounded. As the variances becomes infinitly small the "Y" becomes infinitly large. But is there any interpretation to th "Y". P(X=c)=Some Y Value. I know the area underneath this single point is 0, but does the "Y" value tell us something? Ok forget that the distribtuion is continious, and we have possible X values as our X-Scale, and the frequency as our Y-scale. Should I divide the frequency of the event by the number of observation to get the probability of the event? –  gabriel Oct 20 '12 at 1:46
I have added a little to the post. Hope it helps answer your question. –  André Nicolas Oct 20 '12 at 1:53

To add some commentary, the "bell curve" shape is governed by the PDF, as @AndreNicolas pointed out. However, the actual "y"-value of this curve is itself more or less meaningless. The integral of the PDF $f(x)$ gives the probability that your random variable is less than some value: $P(x < X) = \int_{-\infty}^X f(x)dx$. This is known as the CDF, or cumulative distribution function. By the fundamental theorem of calculus, the PDF is then the derivative of the CDF; that is, the PDF is the derivative of a function that returns a probability. So what is that intuitively? Honestly... it's not really anything. The "units" of the vertical axis in the PDF plot don't lead to anything intuitive; they are meaningful, but only in a derived, mathematical sense.

Some people wish to think that $f(X)$ is the probability that $x = X$, but this is untrue for continuous distributions ($P(x = X) = 0$). However, for the PDF's discrete analog, the Probability Mass Function (PMF), this statement is quite true.

share|improve this answer
But as a pdf shouldn't it be the case the function is always less than 1? I mean the area underneath the function should sum to 1 but if the pdf is a probability function shouldn't that probability be less than or equal to 1? –  gabriel Oct 20 '12 at 1:41
@gabriel The area under the pdf equals $1$. If the pdf value $f(x)$ exceeds $1$ for some and indeed many values of $x$, that is perfectly fine: but $f(x)$ cannot exceed $1$ for all $x$ in an interval $I$ of length exceeding $1$. If the latter condition were to hold, then $$\int_I f(x)\,\mathrm dx=\text{area under pdf in interval}~ I>1$$ in violation of the constraint that the total area is $1$. The value of $f(x)$ is not a probability. The units of $f(x)$ are probability per unit length and you must multiply by length (more generally, find an area) to get a probability. –  Dilip Sarwate Oct 20 '12 at 2:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.