I am looking for something similar to the normal distribution in which there are a mean and a standard deviation representing the amount of variation. However, the value must always be positive, i.e. $P(x \leq 0) = 0$.
$\begingroup$
$\endgroup$
8
-
2$\begingroup$ What about a Gamma or a log-normal distribution ? There is a mean and a standard deviation for these distributions. Do you mean something else by "similar to the normal distribution" ? $\endgroup$– Stéphane LaurentCommented Aug 21, 2015 at 8:51
-
$\begingroup$ That means I want something that has the simibar shape to normal distribution. $\endgroup$– Long ThaiCommented Aug 21, 2015 at 8:57
-
2$\begingroup$ You could transform a symmetric Beta distribution to any interval $[a,b]$ using an affine transformation. $\endgroup$– Stéphane LaurentCommented Aug 21, 2015 at 9:03
-
2$\begingroup$ @FedericoPoloni he means the support, not the density $\endgroup$– MichaelChiricoCommented Aug 21, 2015 at 14:18
-
3$\begingroup$ @LongThai you should also recognize that the strictly positive distribution referred to in the question you link is quite different from the strictly positive distribution you describe--namely, those theorems refer to the density being strictly positive, while it seems you are referring to the support being strictly positive. This is the same as focusing on, for $f:D\rightarrow R$, whether the domain $D$ of $f$ (your question) or the range $R$ of $f$ (linked question) is restricted to be positive. $\endgroup$– MichaelChiricoCommented Aug 21, 2015 at 15:53
|
Show 3 more comments
4 Answers
$\begingroup$
$\endgroup$
You may be interested in this gallery of distributions. In addition to
- the gamma distribution
- the lognormal distribution
- the $\chi^2$ distribution
- and the truncated normal distribution
that have already been brought up, you could check
- the F distribution
- the exponential distribution
- the Weibull distribution
- the power lognormal distribution
All of these have defined means and variances. Pick the one you like best.
answered Aug 21, 2015 at 14:31
$\begingroup$
$\endgroup$
2
What about the truncated normal distribution? Try for example
library(truncnorm)
x <- seq(0,10,by=.01)
plot(x,dtruncnorm(x, a=3, b=Inf, mean = 5, sd = 1),type="l")
This gives
By taking the mean of the underlying normal distribution $\mu$ (mean in the command) larger you can make it look "almost" normal, without a nonzero probability of nonpositive values.
answered Aug 21, 2015 at 8:29
-
$\begingroup$ Thanks for your answer, I think the only issue is that the mean is not exactly the same with the given value. But as you stated, using a larger mean can help. $\endgroup$ Commented Aug 21, 2015 at 9:00
-
$\begingroup$ Indeed. As you cut off probability mass in the left part of the distribution, there is no way the mean can stay exactly the same. $\endgroup$ Commented Aug 21, 2015 at 9:03
$\begingroup$
$\endgroup$
2
There are infinitely many such distributions ...
Consider the family of uniform distributions from $0$ to $N$ (non-inclusive of $0$), where $N$ is an arbitrary integer.
Now choose any one of these, say $X_1 \sim U(0,3)$.
Then the sum of $X_1 + \dots + X_{10}$, where $X_1, \dots, X_{10} \overset{\text{iid}} \sim U(0,3)$ will be approximately normal in shape.
This uniform sum distribution is also known as the Irwin-Hall distribution.
-
2$\begingroup$ In the comments above, OP mentions that he wants "something that has a shape similar to the normal distribution" $\endgroup$ Commented Aug 21, 2015 at 12:00
-
$\begingroup$ good eye. will emend to reflect. $\endgroup$ Commented Aug 21, 2015 at 14:55
$\begingroup$
$\endgroup$
Two suggestions for you:
- The Non-central $\chi^2_1$ distribution which is (can be) obtained by squaring the normal distribution $N(\mu,1)$?
This easily satisfies your relation to normal as it always shares the parameters of some underlying normal. However, it always skews right, so while it may look sort of normal, it will never be.
- The distribution of number of heads from a series $n$ of coin flips (also known as the $Binomial(n,p)$ distribution) looks more and more like the normal distribution as $n$ increases.
The drawback here is that the support is always finite and includes 0; this latter can easily be remedied by just adding 1.
answered Aug 21, 2015 at 14:25