11
$\begingroup$

We can write Bayes' theorem as

$$p(\theta|x) = \frac{f(X|\theta)p(\theta)}{\int_{\theta} f(X|\theta)p(\theta)d\theta}$$

where $p(\theta|x)$ is the posterior, $f(X|\theta)$ is the conditional distribution, and $p(\theta)$ is the prior.

or

$$p(\theta|x) = \frac{L(\theta|x)p(\theta)}{\int_{\theta} L(\theta|x)p(\theta)d\theta}$$

where $p(\theta|x)$ is the posterior, $L(\theta|x)$ is the likelihood function, and $p(\theta)$ is the prior.

My question is

  1. Why is Bayesian analysis done using the likelihood function and not the conditional distribution?
  2. Can you say in words what the difference between the likelihood and conditional distribution is? I know the likelihood is not a probability distribution and $L(\theta|x) \propto f(X|\theta)$.
$\endgroup$
  • 1
    $\begingroup$ There is no difference! The likelihood is the conditional distribution $f(X | \theta)$, well, is proportional to, which is all that matters. $\endgroup$ – kjetil b halvorsen Sep 17 '12 at 2:02
  • $\begingroup$ Prior parameter $\Theta$ has density $p_\Theta(\theta)$. if the realization of $\Theta$ has value $\theta$ while $x$ is the observed value of a random variable $X$, then the value of the likelihood function $L(\theta\mid x)$ is precisely $f(x\mid \theta)$, the value of the conditional density $f_{X\mid\Theta}(x\mid\Theta=\theta)$ of $X$. The difference is that $$\int_{-\infty}^{\infty}f_{X\mid\Theta}(x\mid\Theta=\theta)dx=1$$ for all realizations of $\Theta$. However, as a function of $\theta$ (and fixed $x$), $L(\theta\mid x)$ is not a density: $$\int L(\theta\mid x)d\theta\neq 1$$ $\endgroup$ – Dilip Sarwate Sep 17 '12 at 2:08
10
$\begingroup$

Suppose that you have $X_1,\dots,X_n$ random variables (whose values will be observed in your experiment) that are conditionally independent, given that $\Theta=\theta$, with conditional densities $f_{X_i\mid\Theta}(\,\cdot\mid\theta)$, for $i=1,\dots,n$. This is your (postulated) statistical (conditional) model, and the conditional densities express, for each possible value $\theta$ of the (random) parameter $\Theta$, your uncertainty about the values of the $X_i$'s, before you have access to any real data. With the help of the conditional densities you can, for example, compute conditional probabilities like $$ P\{X_1\in B_1,\dots,X_n\in B_n\mid \Theta=\theta\} = \int_{B_1\times\dots\times B_n} \prod_{i=1}^n f_{X_i\mid\Theta}(x_i\mid\theta)\,dx_1\dots dx_n \, , $$ for each $\theta$.

After you have access to an actual sample $(x_1,\dots,x_n)$ of values (realizations) of the $X_i$'s that have been observed in one run of your experiment, the situation changes: there is no longer uncertainty about the observables $X_1,\dots,X_n$. Suppose that the random $\Theta$ assumes values in some parameter space $\Pi$. Now, you define, for those known (fixed) values $(x_1,\dots,x_n)$ a function $$ L_{x_1,\dots,x_n} : \Pi \to \mathbb{R} \, $$ by $$ L_{x_1,\dots,x_n}(\theta)=\prod_{i=1}^n f_{X_i\mid\Theta}(x_i\mid\theta) \, . $$ Note that $L_{x_1,\dots,x_n}$, known as the "likelihood function" is a function of $\theta$. In this "after you have data" situation, the likelihood $L_{x_1,\dots,x_n}$ contains, for the particular conditional model that we are considering, all the information about the parameter $\Theta$ contained in this particular sample $(x_1,\dots,x_n)$. In fact, it happens that $L_{x_1,\dots,x_n}$ is a sufficient statistic for $\Theta$.

Answering your question, to understand the differences between the concepts of conditional density and likelihood, keep in mind their mathematical definitions (which are clearly different: they are different mathematical objects, with different properties), and also remember that conditional density is a "pre-sample" object/concept, while the likelihood is an "after-sample" one. I hope that all this also help you to answer why Bayesian inference (using your way of putting it, which I don't think is ideal) is done "using the likelihood function and not the conditional distribution": the goal of Bayesian inference is to compute the posterior distribution, and to do so we condition on the observed (known) data.

$\endgroup$
  • $\begingroup$ I think Zen is correct when he says that the likelihood and conditional probability are different. In the likelihood function θ is not a random variable, thus it is different from conditional probability. $\endgroup$ – Martine May 23 '17 at 21:11
1
$\begingroup$

Proportionality is used to simplify analysis

Bayesian analysis is generally done via an even simpler statement of Bayes' theorem, where we work only in terms of proportionality with respect to the parameter of interest. For a standard IID model with sampling density $f(X|\theta)$ we can express this as:

$$p(\theta|\mathbf{x}) \propto L_\mathbf{x}(\theta) \cdot p(\theta) \quad \quad \quad \quad L_\mathbf{x}(\theta) \propto \prod_{i=1}^n f(x_i|\theta).$$

This statement of Bayesian updating works in terms of proportionality with respect to the parameter $\theta$. It uses two proportionality simplifications: one in the use of the likelihood function (proportional to the sampling density) and one in the posterior (proportional to the product of likelihood and prior). Since the posterior is a density function (in the continuous case), the norming rule then sets the multiplicative constant that is required to yield a valid density (i.e., to make it integrate to one).

This method use of proportionality has the advantage of allowing us to ignore any multiplicative elements of the functions that do not depend on the parameter $\theta$. This tends to simplify the problem by allowing us to sweep away unnecessary parts of the mathematics, and get simpler statements of the updating mechanism. This is not a mathematical requirement (since Bayes' rule works in its non-proportional form too), but it makes things simpler for our tiny animal brains.

An applied example: Consider an IID model with observed data $X_1, ..., X_n \sim \text{IID N}(\theta, 1)$. To facilitate our analysis we define the statistics $\bar{x} = \tfrac{1}{n} \sum_{i=1}^n x_i$ and $\bar{\bar{x}} = \tfrac{1}{n} \sum_{i=1}^n x_i^2$, which are the first two sample moments. For this model we have sampling density:

$$\begin{equation} \begin{aligned} f(\mathbf{x}|\theta) = \prod_{i=1}^n f(x_i|\theta) &= \prod_{i=1}^n \text{N}(x_i|\theta,1) \\[6pt] &= \prod_{i=1}^n \frac{1}{\sqrt{2 \pi}} \exp \Big( -\frac{1}{2} (x_i-\theta)^2 \Big) \\[6pt] &= (2 \pi)^{n/2} \exp \Big( -\frac{1}{2} \sum_{i=1}^n (x_i-\theta)^2 \Big). \\[6pt] &= (2 \pi)^{n/2} \exp \Big( -\frac{n}{2} ( \theta^2 - 2\bar{x} \theta + \bar{\bar{x}} ) \Big) \\[6pt] &= (2 \pi)^{n/2} \exp \Big( -\frac{n \bar{\bar{x}}}{2} \Big) \cdot \exp \Big( -\frac{n}{2} ( \theta^2 - 2\bar{x} \theta ) \Big) \\[6pt] \end{aligned} \end{equation}$$

Now, we can work directly with this sampling density if we want to. But notice that the first two terms in this density are multiplicative constants that do not depend on $\theta$. It is annoying to have to keep track of these terms, so let's just get rid of them, so we have the likelihood function:

$$L_\mathbf{x}(\theta) = \exp \Big( -\frac{n}{2} ( \theta^2 - 2\bar{x} \theta ) \Big).$$

That simplifies things a little bit, since we don't have to keep track of an additional term. Now, we could apply Bayes' rule using its full equation-version, including the integral denominator. But again, this requires us to keep track of another annoying multiplicative constant that does not depend on $\theta$ (more annoying because we have to solve an integral to get it). So let's just apply Bayes' rule in its proportional form. Using the conjugate prior $\theta \sim \text{N}(0,\lambda_0)$, with some known precision parameter $\lambda_0>0$, we get the following result (by completing the square):

$$\begin{equation} \begin{aligned} p(\theta|\mathbf{x}) &\propto L_\mathbf{x}(\theta) \cdot p(\theta) \\[10pt] &= \exp \Big( -\frac{n}{2} ( \theta^2 - 2\bar{x} \theta ) \Big) \cdot \text{N}(\theta|0,\lambda_0) \\[6pt] &\propto \exp \Big( -\frac{n}{2} ( \theta^2 - 2\bar{x} \theta ) \Big) \cdot \exp \Big( -\frac{\lambda_0}{2} \theta^2 \Big) \\[6pt] &= \exp \Big( -\frac{1}{2} ( n\theta^2 - 2n\bar{x} \theta + \lambda_0 \theta^2 ) \Big) \\[6pt] &= \exp \Big( -\frac{1}{2} ( (n+\lambda_0) \theta^2 - 2n\bar{x} \theta ) \Big) \\[6pt] &= \exp \Big( -\frac{n+\lambda_0}{2} \Big( \theta^2 - 2 \frac{n\bar{x}}{n+\lambda_0} \theta \Big) \Big) \\[6pt] &\propto \exp \Big( -\frac{n+\lambda_0}{2} \Big( \theta - \frac{n}{n+\lambda_0} \cdot \bar{x} \Big)^2 \Big) \\[6pt] &\propto \text{N}\Big( \theta \Big| \frac{n}{n+\lambda_0} \cdot \bar{x}, n+\lambda_0 \Big). \\[6pt] \end{aligned} \end{equation}$$

So, from this working we can see that the posterior distribution is proportional to a normal density. Since the posterior must be a density, this implies that the posterior is that normal density:

$$p(\theta|\mathbf{x}) = \text{N}\Big( \theta \Big| \frac{n}{n+\lambda_0} \cdot \bar{x}, n+\lambda_0 \Big).$$

Hence, we see that a posteriori the parameter $\theta$ is normally distributed with posterior mean and variance given by:

$$\mathbb{E}(\theta|\mathbf{x}) = \frac{n}{n+\lambda_0} \cdot \bar{x} \quad \quad \quad \quad \mathbb{V}(\theta|\mathbf{x}) = \frac{1}{n+\lambda_0}.$$

Now, the posterior distribution we have derived has a constant of integration out the front of it (which we can find easily by looking up the form of the normal distribution). But notice that we did not have to worry about this multiplicative constant - all our working removed (or brought in) multiplicative constants whenever this simplified the mathematics. The same result can be derived while keeping track of the multiplicative constants, but this is a lot messier.

$\endgroup$
0
$\begingroup$

I think Zen's answer really tells you how conceptually the likelihood function and the joint density of values of random variables differ. Still mathematically as a function of both the x$_i$s and θ they are the same and in that sense the likelihood can be looked at as a probability density. The difference you point to in the formula for the Bayes posterior distribution is just a notational difference. But the subtlety of the difference is nicely explained in Zen's answer.

This issue has come up in other questions discussed on this site regarding the likelihood function. Also other comments by kjetil and Dilip seem to support what I am saying.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.