1. ## recognising uniform distibution

I am going through an example where one needs to find score and information for a random sample of size n from the logistic distribution with a density function

$f_Y(y;\mu)=\frac{e^{y-\mu}}{[1+e^{(e-\mu)}]^2}$

and the score is $s_Y(Y;\mu)=\Sigma_{i=1}^n(-1+2F_Y(y_i;\mu))$

then the explanation goes, 'by recognising that $F_Y(Y;\mu)\sim Unif[0;1]$ we can calculate information... and they use the formula for Var of a Uniform distribution.

Can someone help me to see that $F_Y(Y;\mu)\sim Unif[0;1]$?

2. Hello,

That's a standard property, if Fy has an inverse (it's the case in particular if the rv has a continuous density).

Indeed : $P(F_Y(Y)\leq x)=P(Y\leq F_Y^{-1}(x))=F_Y(F_Y^{-1}(x))=x$, for $x\in[0,1]$ since $F_Y$ is a probability and is hence taking values between 0 and 1.

3. Originally Posted by Volga
I am going through an example where one needs to find score and information for a random sample of size n from the logistic distribution with a density function

$f_Y(y;\mu)=\frac{e^{y-\mu}}{[1+e^{(e-\mu)}]^2}$

and the score is $s_Y(Y;\mu)=\Sigma_{i=1}^n(-1+2F_Y(y_i;\mu))$

then the explanation goes, 'by recognising that $F_Y(Y;\mu)\sim Unif[0;1]$ we can calculate information... and they use the formula for Var of a Uniform distribution.

Can someone help me to see that $F_Y(Y;\mu)\sim Unif[0;1]$?
There are a number of ways of looking at this, one is the usual transformation method where if $U$ has pdf $f_U(u)$ we determine the distribution of $V=h(U)$ where $h(.)$ is an increasing function.

$f_V(v)=\dfrac{f_U(u)}{h'(u)}$

But if $h(u)=F_U(u)$ then $h'(u)=f_U(u)$ so $f_V(v)=1$ in the range of $h(u)$ (which is $[0,1]$).

CB

4. Nice one, I didn't know that, thank you!