A small doubt on the notation for the induced likelihood function and invariance properties of maximum likelihood estimators - It says in a textbook I am reading that:

Let $X = {X1, X2,..., Xn}$ be a random sample parameterized by $θ$. Suppose $φ = g(θ)$ where $g : Θ → Φ$. The induced likelihood function of $φ$ given an observed sample $x$ is:

$L'$X$(φ; x) = sup ${θ:g(θ)=φ}$ L$X$(θ; x)$

The book then proceeds to prove the invariance of MLEs by proving why $L'$X$(\hat{φ}; x)$ = $L'$X$(g(\hat{θ}); x)$.

So my questions are:

i) Is the apostrophe in $L'$X$(φ; x)$ supposed to denote the differential of $L'$X or just a variant of $L$X?

ii) In the first equation, shouldn't it be $L'$X$(\hat{φ}; x)$ instead of $L'$X$(φ; x)$? Why is $sup ${θ:g(θ)=φ}$ L$X$(θ; x)$ equal to the likelihood of $x$ for just any $φ$?

iii) What is the purpose of proving the second equation to support the first one? Haven't we already defined $φ$ as $g(θ)$?

Thanks in advance for your help!


1 Answer 1


i) Yes the apostrophe is here to emphasize that these are two differents functions (that may return differents values).

Take for example $X \sim N(\theta, 1)$ and $g(\theta) = \theta-1$ then $$ L_X(1,x) = \frac{1}{\sqrt{2 \pi}} \exp \left ( -\frac{(x-1)^2}{2} \right ) $$ while $$ L'_X(1,x) = \frac{1}{\sqrt{2 \pi}} \exp \left ( -\frac{(x-2)^2}{2} \right ) $$

thus $L'_X(1,x) \neq L_X(1,x)$.

ii) In the first equation the function $L'_X$ is defined w.r.t the parameter $\phi$ while $\hat \phi$ is the value that maximizes $L'_X$ (if such value exists).

For the question

Why is sup{θ:g(θ)=φ}LX(θ;x) equal to the likelihood of x for just any φ?

This is not a stated equality it is a definition, meaning that $L'_X(\phi, x)$ is defined that way. This in order to avoid the cases where two distincts values of $\theta$ lead to the same $\phi$: i.e if we have $g(\theta_1) = g(\theta_2) = \phi$ how do we define the likelihood regarding the "transformed" parameter $\phi$? A proposal is to take $\sup_{\theta : g(\theta) = \phi} L_X(\theta,x)$.

This is needed when for example the transform $\theta \mapsto \phi$ is not bijective. This is not the case when $g$ is invertible since there will be only one value $\theta$ for which $g(\theta) = \phi$. In that case $L'_X(\phi, x) = L_X(g^{-1}(\phi),x)$.

iii) Now we want to find the value $\hat \phi$ such that $L'_X(\hat \phi,x)$ reaches its maximum. That is $$ \hat \phi := \text{arg}\max_\phi L'_X(\phi, x) = \text{arg}\max_\phi \sup_{\theta : g(\theta) = \phi} L_X(\theta, x) $$

If $g$ is invertible this reduces to find $$ \hat \phi := \text{arg}\max_\phi L'_X(\phi, x) = \text{arg}\max_\phi L_X(g^{-1}(\phi), x) $$

In that case, since the MLE $\hat \theta$ is defined as $$ \hat \theta = \text{arg}\max_\theta L_X(\theta, x) $$

we have $g^{-1}(\hat \phi ) = \hat \theta$ and thus $\hat \phi = g(\hat \theta)$.

The invariance of MLE states that this holds even when $g$ is not invertible anymore, i.e the value that maximizes $L'_X(\phi,x)$ is $g(\hat \theta)$ where $\hat \theta$ is the value that maximizes $L_X(\theta,x)$. This is a property of the induced likelihood that holds because of the way it has been defined.


We have $$ L'_X( \hat \phi , x) = \sup_\phi L'(\phi , x) = \sup_\phi \sup_{\theta : g(\theta) = \phi} L_X(\theta , x) $$ One "property" of the maximization process of a function is that

$$ \sup_y \sup_{x \in R_y} f(x) = \sup_{x \in \cup_y R_y} f(x) $$

Applying this yield

$$ \sup_\phi \sup_{\theta : g(\theta) = \phi} L_X(\theta , x) = \sup_\theta L_X(\theta, x) = L_X(\hat \theta, x) $$

Thus we have $$ L'_X(\hat \phi, x) = L_X(\hat \theta, x) \qquad (1) $$

Finally we want to show that $\hat \phi = g(\hat \theta)$.

Since $\hat \theta$ is such that $\sup_\theta L_X(\theta, x) = L(\hat \theta ,x )$ , for any function $f$ we have,

$$ L_X(\hat \theta , x) = \sup_{\theta : f(\theta) = f(\hat \theta)} L_X(\theta,x) $$

since $\hat \theta \in \left \{ \theta : f(\theta ) = f(\hat \theta ) \right \}$.

In particular if $f=g$ we have

\begin{align*} L_X(\hat \theta , x) &= \sup_{ \theta : g(\theta ) = g(\hat \theta) } L_X(\theta, x) \\ &= L'_X(g(\hat \theta) , x) \qquad (2) \end{align*}

And finally combining $(1)$ and $(2)$ shows that the MLE for $L'_X$, $\hat \phi$, is $g(\hat \theta)$.

  • $\begingroup$ Hi @winperikle, thank you so much for the clarification. I know fully understand (i) and (ii), but am still a little unsure about (iii). The textbook I am reading goes on to actually prove the invariance when g is no longer invertible, but in a rather complex way. It says that Lx(θ^;x) = sup {θ∈Θ} Lx(θ;x) = sup {θ:g(θ)=g(θ^)} Lx(θ; x) = L'x(g(θ^); x). $\endgroup$ Mar 25, 2021 at 4:10
  • $\begingroup$ I know this a bit unclear (but it's not allowing me to use Latex in the comments). Basically, it says that the likelihood of the estimator ϕ is equal to the likelihood of the estimator θ. This likelihood is equal to the supremum of the likelihood of just θ (w.r.t. to θ∈Θ). And this is, in turn, equal to the the same likelihood, but now w.r.t. to θ:g(θ)=g(θ^). And finally, this equals to the likelihood of the estimator g(θ^). $\endgroup$ Mar 25, 2021 at 4:10
  • $\begingroup$ Do you have any idea of why changing the w.r.t. of the likelihood of just θ changed the equation to the likelihood of the estimator g(θ^)? $\endgroup$ Mar 25, 2021 at 4:10
  • $\begingroup$ @Academic005 I have edited my answer, I hope that helps a bit. $\endgroup$
    – periwinkle
    Mar 25, 2021 at 13:57

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