# How can the density of a truncated normal distribution be greater than one? [duplicate]

According to the info in the following locations:

The probability density function for a left-truncated normal distribution should be:

$f(x) = {{1 \over \sigma}\phi({{x-\mu} \over \sigma})\over{1-\Phi({{a-\mu} \over \sigma})}}$

for all x >= a

However, I'm getting some weird results when I do that.

if I take $\sigma$ = 1.0 and $\mu$=0.0 and $a$=1.0, that gives:

$f(x) = {\phi(x) \over {1-\Phi(1)}}$

If I then plugin a number near the truncation point (say 1.1), I get:

$f(x) \approx {0.212 \over {0.159}} \approx 1.3$

Clearly that's wrong (the PDF should be in the range (0.0, 1.0) for all X).

Except, everywhere I look leads me to the definition of f above, so I can't figure out what I'm doing wrong.

Clearly though, I've gotten something wrong.

I know this is an embarrassingly simple question, but I'd appreciate any help you might be able to give. Do you know what I'm doing wrong?

• The PDF does not have to be in the range (0, 1), it just needs to integrate to 1 over its domain. For example, the continuous uniform distribution U(0, 1/2) has PDF $f(x)=2$.
– MLaz
Aug 19, 2014 at 5:18
• A common mistake. Density is not probability. The issue here effectively duplicates this or this and a number of others. Aug 19, 2014 at 5:26
• It's the area under a range within its support that will be between 0 and 1.
– Luca
Aug 19, 2014 at 7:45
• Even an untruncated normal density function can have value greater than $1$. Consider, for example, the value of the density at $0$ for a normal random variable with mean $0$ and standard deviation smaller than $\frac{1}{\sqrt{2\pi}}$. Aug 19, 2014 at 12:37