Recitation: The PDF of the Absolute Value of X

What this lecture covers

This short recitation problem shows how to find the PDF of Y = |X| given the PDF of X, using two concrete examples before generalizing to a formula. It builds directly on the derived-distributions method, applied to the specific and commonly useful case of an absolute value.

The video reasons first from the discrete case, where a value of |X| can come from two different values of X, then transfers that reasoning to continuous densities.

Key ideas

• Two contributions per value: just as in the discrete case where P(|X|=v) combines the probability at X=v and X=-v, the density of Y=|X| at a point y combines the density of X at y and at -y.
• Fold-and-stack visualization: graphically, taking the PDF of X, flipping the negative-side portion over the y-axis, and stacking it on top of the positive-side portion gives the PDF of |X|.
• One-sided densities need no folding: if X is already always non-negative, then |X| = X and the PDF is unchanged.
• General formula: the PDF of Y=|X| equals the PDF of X at y plus the PDF of X at -y, for y >= 0.

Walkthrough

A two-piece uniform example (0:01)

For X uniform with density 1/3 on [-2,1], the video reasons that the density of Y=|X| at a given value has two contributions, one from the positive and one from the negative side of X's range. Folding the negative portion onto the positive portion and stacking gives a two-level step function: density 2/3 for y between 0 and 1 (where both sides overlap), and density 1/3 for y between 1 and 2 (where only the negative side contributes).

A one-sided exponential example (4:03)

For X exponential with rate 2 (already non-negative), there is nothing on the negative side to fold, so |X| has exactly the same PDF as X.

The general formula (6:11)

Generalizing both examples, the PDF of Y=|X| at y is the PDF of X at y plus the PDF of X at -y, mirroring the discrete formula that adds P(X=y) and P(X=-y).

Before you watch

• Review the general method for finding the distribution of a function of a random variable.
• It helps to first think through the discrete analog of taking an absolute value before moving to densities.

Check your understanding

1. Why does the density of |X| at a point y generally involve two terms rather than one?
2. Why does the PDF of |X| equal the PDF of X unchanged when X is already non-negative?
3. In the uniform example, why is the density of |X| higher between 0 and 1 than between 1 and 2?