Density Scaling After Standardizing a Random Variable

# Problem with obtaining densities


# Problem with obtaining densities












For my research I need to obtain a series of densities, however, I am encountering some problems.

The first problem is perhaps very simple, but the answer eludes me. Let's say I have an observation from a time series, $x=0.001$ together with estimated mean $\mu=0.0001$ and standard deviation $\sigma=0.4$. Using R:

```
dnorm(0.001, 0.0001, 0.4)
z=(0.001-0.0001)/0.4
dnorm(z, 0, 1)
```

where z is the standardized variable. Why are the result different?

The second issue is connected to Hansen's Skew-t distribution. Let's add the skew $\lambda=0.1$ and shape $\eta=5$ parameters. When plugging in the all the parameters into the pdf (I take z which takes $\mu$ and $\sigma$ into account) I obtain $0.4832$, but when I use the `sgt` package in R:

```
dsgt(0.001, 0.0001, 0.4, 0.1, p = 2, 5, mean.cent=TRUE, var.adj=TRUE)
```

Iobtain $1.075749$. Inputting the standardized variable z and $\mu = 0$ and $\sigma = 1$ into the above yields $0.5425881$.

Can you please explain the reason why the values are different?

## Answer by Malick (score 3, accepted)

https://quant.stackexchange.com/a/22492

You know that :

$X \sim N(\mu,\sigma^2)$.

$Z = \large\frac{X-\mu}{\sigma}$.

$\text{Var}(Z) = \large\frac{1}{\sigma^2}\text{Var}(X) = \large\frac{1}{\sigma^2}\sigma^2 = 1$.

So that $Z \sim N(0,1)$.

> However note that the pdf evaluated for X and Z have different domains.

The following figure illustrate it :

- $X$ is plotted in a) and $Z$ in b)

- Their respective normal pdf are plotted in c) and d). Note that their integrals equals 1.

- In e) I applied the pdf of Z on the (wrong) original domain , notice that the integral is different of 1.

- To obtain the correct pdf on the original domain we need to scale the pdf(Z) by ($1/ \sigma$), this is done in f). The integral is ok and equal to 1.

The vertical line is evaluated at X= 1.5, you see that the density differs accordingly the domain.

So in your example you also need to scale the density by ($1/\sigma$) :

`(1/sigma)*dnorm(z, 0, 1) = dnorm(0.001, 0.0001, 0.4)`

To summarize when you use a normal transformation you need to scale the density by ($1/ \sigma$) to get the correct density in the original domain.

> You can find the matlab code used in this answer here.

I think you second issue and some of your others questions are related to this problem.

## Answer by owner (score 1)

https://quant.stackexchange.com/a/22493

Have a look at `?dnorm`, and rather use the `standardized value` as argument in your function, in addition to `mean` and `sd`:

`a_<-dnorm((0.001-0.0001)/0.4, mean=0, sd=1)`

Hope it helps

[EDIT] Likewise from `?dsgt`

`st<-(0.001-0.0001)/0.4`

`skewt<-dsgt(st, mu=0, sigma=1, lambda=0.1, p = 2, q=5, mean.cent=TRUE, var.adj=TRUE)`

results in `skewt=0.4302996` (close to your value)

For an explanation of the differences between `Hansen(1994)` and the `recent references` used in `dsgt` maybe it's worth posting a thread to Cross Validated (see link for related discussions)

Hope it helps