I was just wondering how I would go about creating a likelihood function if I have a $N( \theta,1)$ distribution and know $x(n)$ the maximum of n observations, but not the actual observations themselves. I understand how to create a CDF for the maximum order statistic (i.e. $F(x)^n $), but the normal distribution does not have a closed form CDF.
Am I able to find the likelihood function by taking the derivative of $F(x)^n $ (in integral form) and then evaluate the derivative at $x(n)$? Or is my reasoning flawed/or is there an easier way to do this?
Any help would be appreciated. Thank you!
We have the functions $\Phi()$ and $\phi()$, standard notation for the cdf and pdf respectively of a $N(0,1)$ distribution. Or if you want to use $F$ and $f$, that also makes no difference as long as you define them. The fact that the cdf does not have a closed form does not stop us from writing down the likelihood at least.
Assuming you have i.i.d observations $X_1,\ldots,X_n\sim N(\theta,1)$, the distribution function of $T=\max\limits_{1\le k\le n}X_k$ is
$$P(T\le t)=(P(X_1\le t))^n=(\Phi(t-\theta))^n\quad,\,t\in\mathbb R$$
So the pdf of $T$ is
$$f_{T}(t)=n(\Phi(t-\theta))^{n-1}\phi(t-\theta)\quad,\,t\in\mathbb R$$
The likelihood function given $t \,(\in\mathbb R)$, the observed value of $T$, is therefore
$$L(\theta\mid t)=n(\Phi(t-\theta))^{n-1}\phi(t-\theta)\quad,\,\theta\in\mathbb R$$