You start with a model, usually given as a family of densities (or probability mass functions). We write this as $$ Y \sim f(y;\theta) ~~\text{ for $\theta \in \Theta$ } $$ where $\Theta$ is the parameter space. Here, the meaning is that $\theta$ is an unknown parameter. For each fixed $\theta$ this gives a probability distribution for data $Y$. Now, the likelihood function is a function of $\theta$ for fixed $y$, there is nothing in the setup which coud imply this to be a density function!, sometimes, by accident it "is" so (because by accident it integrates to one), but that accident has nothing to do with its logical status. Sometimes (often) it is not. An instructive example is $$ Y \sim U(0, \theta) \qquad \theta > 0 $$ That is, a uniform distribution. If we now observes $Y=1$, then the likelihood function becomes $$ L(\theta) = \frac{1}{\theta} ~~ \text{for $\theta > 1$} $$ which has total integral $\int_{1}^{\infty} \frac{1}{\theta}\, d\theta $ which do not converge (that is, is $\infty$) so clearly is not a probability density neither can be normalized to be.

You start with a model, usually given as a family of densities (or probability mass functions). We write this as $$ Y \sim f(y;\theta) ~~\text{ for $\theta \in \Theta$ } $$ where $\Theta$ is the parameter space. Here, the meaning is that $\theta$ is an unknown parameter. For each fixed $\theta$ this gives a probability distribution for data $Y$. Now, the likelihood function is a function of $\theta$ for fixed $y$, there is nothing in the setup which could imply this to be a density function!, sometimes, by accident it "is" so (because by accident it integrates to one), but that accident has nothing to do with its logical status. Sometimes (often) it is not. An instructive example is $$ Y \sim U(0, \theta) \qquad \theta > 0 $$ That is, a uniform distribution. If we now observes $Y=1$, then the likelihood function becomes $$ L(\theta) = \frac{1}{\theta} ~~ \text{for $\theta > 1$} $$ which has total integral $\int_{1}^{\infty} \frac{1}{\theta}\, d\theta $ which do not converge (that is, is $\infty$) so clearly is not a probability density neither can be normalized to be.

You start with a model, usually given as a family of densities (oor probability mass functions). We write this as $$ Y \sim f(y;\theta) ~~\text{ for $\theta \in \Theta$ } $$ where $\Theta$ is the sample space. Here, the meaning is that $\theta$ is an unknown parameter. For each fixed $\theta$ this gives a probability distribution for data $Y$. Now, the likelihood function is a function of $\theta$ for fixed $y$, there is nothing in the setup which coud imply this to be a density function!, sometimes, by accident it "is" so (because by accident it integrates to one), but that accident has nothing to do with its logical status. Sometimes (often) it is not. An instructive example is $$ Y \sim U(0, \theta) \qquad \theta > 0 $$ That is, a uniform distribution. If we now observes $Y=1$, then the likelihood function becomes $$ L(\theta) = \frac{1}{\theta} ~~ \text{for $\theta > 1$} $$ which has total integral $\int_{1}^{\infty} \frac{1}{\theta}\, d\theta $ which do not converge (that is, is $\infty$) so clearly is not a probability density neither can be normalized to be.

You start with a model, usually given as a family of densities (or probability mass functions). We write this as $$ Y \sim f(y;\theta) ~~\text{ for $\theta \in \Theta$ } $$ where $\Theta$ is the parameter space. Here, the meaning is that $\theta$ is an unknown parameter. For each fixed $\theta$ this gives a probability distribution for data $Y$. Now, the likelihood function is a function of $\theta$ for fixed $y$, there is nothing in the setup which coud imply this to be a density function!, sometimes, by accident it "is" so (because by accident it integrates to one), but that accident has nothing to do with its logical status. Sometimes (often) it is not. An instructive example is $$ Y \sim U(0, \theta) \qquad \theta > 0 $$ That is, a uniform distribution. If we now observes $Y=1$, then the likelihood function becomes $$ L(\theta) = \frac{1}{\theta} ~~ \text{for $\theta > 1$} $$ which has total integral $\int_{1}^{\infty} \frac{1}{\theta}\, d\theta $ which do not converge (that is, is $\infty$) so clearly is not a probability density neither can be normalized to be.