# What is the jacobian factor?

The excercise from the book

I am solving problem 1.4 of the famous book Pattern recognition and machine learning of Bishop. The idea of the exercise is that in a simple function $$f(x)$$ the maximum is the same if we apply a transformation $$x = g(y)$$. However, in the case of a probability density, this does not hold anymore. I have solved the exercise, but Bishop says this happens because of the Jacobian factor and I do not understand what this means.

Could someone help me with this concept?

Welcome to MSE! Conceptually the idea is the following: if $$f$$ is a probability density function, it satisfies certain properties, like $$f \geq 0$$ and $$\int f(x) d x = 1.$$ If we look at a transformation $$f(g(y))$$, firstly the second property might be not true anymore. Hence, $$f(g(\cdot))$$ is likely no probability density function. Secondly, what we roughly try to describe with $$f(g(\cdot))$$ is the probability distribution of a random variable $$Y$$ that is given such that $$g(Y)$$ follows the distribution represented by $$f$$. Say $$g$$ is invertible and sufficiently smooth. The distribution of $$Y$$ is given by $$P(Y \in A) = P(g(Y) \in g(A)) = \int_{g(A)} f(x) dx.$$ Which is not very useful in practice, as this integral is on sets of the form $$g(A)$$. According to the integration by substitution formula, we can compute a probability density function $$h$$ such that $$P(Y \in A) = \int_A h(y)dy.$$ Here, $$h = f(g(\cdot))\det J_g(\cdot),$$ where $$J_g$$ is the Jacobian of $$g$$. (Bishop gives this formula in (1.27)).