Recitation: A Derived Distribution Example

What this lecture covers

This short recitation problem applies the derived-distribution method to a specific case: X is a standard normal random variable, and Y is defined piecewise as -X when X is negative and sqrt(X) when X is positive. The goal is to find the PDF of Y.

It follows directly from the lecture's introduction of derived distributions and gives a compact worked example of the four-step process: write the CDF of Y, translate to an event about X, express it using the CDF of X, and differentiate.

Key ideas

• Four-step recipe: write P(Y <= y), rewrite it as an equivalent event about X using the relationship Y = g(X), express that as a difference of CDF values of X, then differentiate to get the PDF of Y.
• Piecewise functions split the translation step: because Y is defined differently for negative and positive X, the event Y <= y corresponds to X lying between -y and y^2.
• Differentiating a range needs the chain rule twice: since both endpoints of the X-range depend on y, differentiating the resulting expression produces two terms, one from each endpoint.
• Range restrictions carry through: because Y is built from -X and sqrt(X), it can never be negative, so the derived PDF is only valid for y > 0 and zero elsewhere.

Walkthrough

Setting up the four-step method (0:00)

X is standard normal, with a known density. Y equals -X for negative X and sqrt(X) for positive X. Writing P(Y <= y) and translating it into a statement about X shows that Y <= y is equivalent to X being between -y and y^2.

Differentiating to get the PDF (5:11)

Rewriting the probability as a difference of two CDF values of X and differentiating with the chain rule (accounting for both endpoints depending on y) produces the PDF of Y as a combination of the normal density evaluated at y^2 and at -y, each with its own chain-rule factor. The final density is valid only for y > 0, since Y can never take negative values.

Before you watch

• Review the two-step (or four-step) derived-distribution procedure from the lecture on continuous Bayes' rule and derived distributions.
• Be comfortable differentiating a CDF-based expression using the chain rule when the limits of integration depend on the target variable.

Check your understanding

1. Why does the event Y <= y translate into X being between -y and y^2 for this particular g?
2. Why does differentiating this expression require the chain rule at two separate points?
3. Why is the resulting PDF of Y only valid for y > 0?