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Probability · Lecture 38 of 76 · 9:30

Recitation: A Derived Distribution Example

Study guide

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

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

Vocabulary

piecewise function (phrase): A function defined by different formulas for different ranges of input.
Y is a piecewise function, defined differently for negative and positive X.
translate (verb): To rewrite one statement in an equivalent, different form.
We translate the event about Y into an event about X.
endpoint (noun): The value marking the start or end of a range.
Both endpoints of the range depend on y.
chain rule (phrase): A calculus rule for differentiating a function made of nested functions.
We apply the chain rule twice, once for each endpoint.
range restriction (phrase): A limit on which values a variable is allowed to take.
The range restriction means the PDF is zero for negative y.
valid (adjective): Correct and applicable under the stated conditions.
The formula is only valid for y greater than 0.
standard normal (phrase): A normal distribution with mean 0 and variance 1.
X is given as a standard normal random variable.
probability density function (PDF) (phrase): A function describing how likely a continuous variable is near each value.
The goal is to find the PDF of Y.
cumulative distribution function (CDF) (phrase): A function giving the probability that a variable is at or below a value.
We express the probability using the CDF of X.
differentiate (verb): To take the derivative of a function.
We differentiate the CDF expression to get the PDF.
express (verb): To write something in a particular mathematical form.
The probability is expressed as a difference of two CDF values.
equivalent (adjective): Having exactly the same meaning or effect.
The event about Y is rewritten as an equivalent event about X.
negative (adjective): Less than zero.
Y equals negative X when X is negative.
positive (adjective): Greater than zero.
Y equals the square root of X when X is positive.
compact (adjective): Expressed briefly, without unnecessary extra parts.
This is a compact worked example of the four-step process.
worked example (phrase): A fully solved example that demonstrates a method step by step.
This recitation gives a worked example of the derived-distribution method.
combination (noun): Something made by joining two or more parts together.
The final PDF is a combination of two density terms.
factor (noun): One part of a multiplication that contributes to the final result.
Each term has its own chain-rule factor.
evaluate (verb): To calculate the value of an expression at a specific point.
The density is evaluated at y squared and at negative y.
specific (adjective): Particular and clearly defined, not general.
This is a specific case of the general derived-distribution method.
correspond to (phrase): To match up with or represent the same thing as another item.
The event Y <= y corresponds to X lying in a certain range.
lie between (phrase): To have a value located within a certain range.
X must lie between negative y and y squared.
define (verb): To state exactly what something means or how it works.
Y is defined piecewise, differently for negative and positive X.
relationship (noun): A connection between two variables or quantities.
The relationship Y = g(X) links Y directly to X.
difference (noun): The result of subtracting one value from another.
The probability is written as a difference of two CDF values.
term (noun): One separate part of a mathematical expression.
Differentiating produces two separate terms in the final answer.

Chapters

0:00 <Untitled Chapter 1>
0:32 To Write Out the Cdf
6:04 Differentiate
6:20 Chain Rule

From the YouTube description

MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
View the complete course: http://ocw.mit.edu/6-041SCF13
Instructor: Jimmy Li

License: Creative Commons BY-NC-SA
More information at http://ocw.mit.edu/terms
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