Re: [R] Truncated normal distribution

From: Ben Bolker <bolker_at_ufl.edu>
Date: Sun, 16 Dec 2007 17:14:22 -0800 (PST)

ravirrrr wrote:
>
> I have the following code, where we need to solve for mu and sigma, when
> we have mut and sdt. Can you suggest how to use a solve function in R to
> do that? I am new to R and am not sure how to go from defining the
> functions, to solving for them.
>
> Thanks
>
>
> truncated <- function(x)
> {
>
> mu=x[1];
> sigma=x[2];
>
>
> f <- function(x) (1/(sigma*sqrt(2*pi)))*exp(-(x-mu)^2/(2*sigma^2));
>
> pdf.fun <- function(x) x*f(x);
>
> sd.fun <- function(x) (x)^2*f(x);
>
> st=integrate(sd.fun,lower=-Inf,upper=1)\$value;
>
> a=integrate(pdf.fun,lower=-Inf,upper=1)\$value;
>
> a1=integrate(f,lower=-Inf,upper=1)\$value;
>
> mut <- a/a1;
> sdt <- sqrt((st/a1)-(a/a1)^2);
>
> }
>

truncated <- function(x)
{
mu <- x[1]
sigma <- x[2]
pdf.fun <- function(x) x*dnorm(x,mu,sigma)

```  sd.fun <- function(x) x^2*dnorm(x,mu,sigma)
st <- integrate(sd.fun,lower=-Inf,upper=1)\$value;
a <- integrate(pdf.fun,lower=-Inf,upper=1)\$value;
```
a1 <- pnorm(1,mu,sigma)
mut <- a/a1;
sdt <- sqrt((st/a1)-(a/a1)^2)
c(mut,sdt)
}

truncated(c(0,1)) ## sensible: mean <0, sd<1 truncated(c(0,0.1)) ## sensible: approx 0,1 ## trouble for small values
truncated(c(0,0.001))
truncated(c(0,0.0001))

optfun <- function(p,target=c(0,1)) {
sum((truncated(p)-target)^2)
}

target <- c(mu=-0.5,sd=2)
fit1 <- optim(fn=optfun,

```      par=c(-0.5,2),
target=target)
```

fit1

truncated(fit1\$par) ## didn't succeed

## let's do something easier -- can we
## work backward to a known value?

t1 <- truncated(c(0,1))

optim(fn=optfun,

```      par=c(0,1),
target=t1)

optim(fn=optfun,
par=c(0.5,2),
target=t1)

```
```--
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