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I have one conceptual problem about a standard normal variate (SNV), $X$.

If $X \sim N(0, 1^2)$, then the probability density function (pdf) is given as

$f(x) = \frac{1}{\sqrt{2\pi}} \exp(-\frac{1}{2} x^2);\quad -\infty<x<\infty$

My Problem :

I understand $f(x)$ as defined above gives the probability that SNV takes the value $X = x$. Thus, $P(X=0)=f(0)$, and $f(0)= 0.398942$ which I understand is very high and it should be high only as mostly all values assumed by $X$ will cluster around $0$.

Similarly, $f(-1) = f(1) = 0.241971$ and $f(-3) = f(3) = 0.004432$

However, as I understand it, since $X$ is a continuous variable, $P(X = 0)$ should be $1 / \infty = 0$.

I am very confused now about my understanding about my concept of pdf. Does the bell shaped curve gives me plot of values assumed by random variable $X$ against the corresponding probability that $(X =x)$. I am not able to understand if my thinking process is right or wrong.

Katherine Gobin
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    Density is not probability. For a continuous variable, the probability of taking any specific value is effectively $f(x)dx$ not $f(x)$. (Consider a normal with mean 0 and standard deviation 0.1; what's $f(0)$ now?). This is addressed in numerous posts. In your situation above, a probability like $P(a – Glen_b Sep 23 '13 at 07:27
  • Many many thanks for making me realize that probability is f(x)dx and f(x). Though I may take some time to understand the concept thoroughly, but your explanation has really helped me to clear my doubts. Thanks again. Regards - Katherine – Katherine Gobin Sep 23 '13 at 07:52

1 Answers1

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@Glen_b's post already answers the question perfectly.

Useful interpretation of $f$: If the proportion $\frac{f(a)}{f(b)} = c$, this tells you that values around $a$ are $c$ times as frequent than values around $b$.

COOLSerdash
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Michael M
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