McGill MATH 323: probability tutoring

Use the central limit theorem for a sample mean

The check above finds the spread of an average. The central limit theorem (CLT) adds a limiting distribution that can help approximate probabilities.

Revisit the sample-mean variance check

Choose the quantity and check the model

1. Name what is random

   Let $X_1,\ldots,X_n$ be independent, identically distributed observations with common mean $\mu$ and finite, positive variance $\sigma^2$. The positive integer $n$ is the sample size, $i$ labels an observation, and $\sigma>0$ is the population standard deviation. The random quantity here is the sample mean $\overline X_n$, not one observation. Its standard deviation, also called its standard error, is $\sigma/\sqrt n$.

2. Standardize the mean

   Define the dimensionless standardized mean by $$Z_n=\frac{\overline X_n-\mu}{\sigma/\sqrt n}.$$ Under these assumptions, the CLT says that the distribution of $Z_n$ approaches the standard normal distribution as $n$ grows. Write $Z\sim N(0,1)$ for a standard normal variable, with mean zero and variance one, and $\Phi(z)=P(Z\le z)$ for its cumulative probability at a dimensionless cutoff $z$.

3. Separate approximate from exact normality

   For a nonnormal population, a finite-sample normal calculation is an approximation. Its accuracy depends on the population’s shape, sample size and event, especially in the tails. There is no universal “$n\ge30$” guarantee. If the independent observations instead have a common normal distribution, their sample mean is exactly normal for every $n$, and $Z_n$ is exactly standard normal.

4. Keep the CLT distinct from the weak law

   The weak law says that, for each fixed tolerance $\varepsilon>0$, the probability $P(|\overline X_n-\mu|\ge\varepsilon)$ tends to zero. The tolerance has the same units as an observation. The CLT describes the limiting shape of the standardized fluctuations $Z_n$. The weak law alone supplies neither a normal approximation nor an exact finite-sample probability.

Worked approximation: an average duration

Use this hypothetical model: each duration $X_i$ is uniform between 0 and 12 minutes, and the durations are independent. All $X_i$, $\overline X_n$, $\mu$ and $\sigma$ are measured in minutes; variance is in squared minutes. Thus $\mu=6$, $\sigma^2=12$ and $\sigma=\sqrt{12}$. For $n=48$, approximate $P(\overline X_{48}>6.5)$: the chance the average exceeds 6.5 minutes.

The population is bounded and symmetric, with finite variance, so this is a reasonable practice setting for a CLT approximation. The individual durations are uniform, not normal; the answer must still be labelled approximate.

Original selected-topic practice, not an official McGill assessment or a complete MATH 323 review. Use your section’s current outline to confirm the methods expected.

Transform a random variable with its CDF

A transformation changes the values a random variable can take, but probabilities still come from the original model. Start with the event $g(X)\le y$, not a memorized density formula. This lesson develops the cumulative distribution function (CDF) method, including the nonmonotone transformation $Y=X^2$.

Prerequisites: integrals of one-variable densities, inequalities, inverse functions and the chain rule. All numerical models below use dimensionless variables. The examples and exercises are original practice, not official assessment questions.

What you will be able to do

• Find the attainable values of a transformed variable before calculating.
• Translate a CDF event into intervals for the original variable.
• Handle a decreasing transformation and a transformation with two inverse branches.
• Check normalization and recognize when a transformed distribution has a point mass.

The support-first method

1. Draw the input and output ranges

   Let $X$ have density $f_X$ on a set $S$, and define $Y=g(X)$. First find the range of $g$ on $S$. A positive input need not produce a positive output; a nonlinear function can turn around inside the support. Include boundary values when describing the closure of the support, even when their individual probabilities are zero.

2. Pull the CDF event back to the input

   For any real cutoff $y$, the CDF is $$F_Y(y)=P(Y\le y)=P(g(X)\le y).$$ Define the input set $A_y=\{x\in S:g(x)\le y\}$. Then $$F_Y(y)=\int_{A_y} f_X(x)\,dx.$$ The set $A_y$ may be empty, one interval, several intervals or all of $S$. A sketch of $g$ helps, but the inequality and support intersection determine the actual integration limits.

3. Separate the cases before differentiating

   For cutoffs below the output range, the CDF is zero; at or above its greatest value, if one exists, the CDF is one. Within the range, locate every cutoff where an inverse branch meets an endpoint of $S$. These are possible breakpoints of the formula. Check that $F_Y$ is nondecreasing and right-continuous, with limits zero and one at the two infinities.

4. Differentiate only where a density describes the law

   On smooth portions of an absolutely continuous distribution, $f_Y(y)=F_Y^{\prime}(y)$. If $g$ is strictly monotone and differentiable, its inverse is differentiable, and $g^{\prime}$ is nonzero on the relevant interval, then $$f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|.$$ The absolute value keeps the density nonnegative. This expression is zero outside the transformed support.

   For finitely many smooth monotone branches, add the contributions from every inverse value that lies in the input support: $$f_Y(y)=\sum_{x:g(x)=y}\frac{f_X(x)}{|g^{\prime}(x)|}.$$ Use this only away from critical values where the displayed denominators vanish. A density may be unbounded near such a value and still integrate to one. The CDF method remains the safer starting point.

5. Check for collapsed intervals and point masses

   A continuous input does not guarantee a continuous output distribution. If $g$ maps an interval of positive input probability to one value $c$, then $P(Y=c)>0$. The CDF jumps at $c$; ordinary differentiation misses that mass. A fully correct answer must state any point masses as well as the continuous part. Exercise 4 makes this distinction explicit.

Two shortcuts that need care

1. A decreasing map reverses the event

   For a continuous $X$ and a strictly decreasing $g$, at an interior output value, $$F_Y(y)=1-F_X(g^{-1}(y)).$$ The event is $X\ge g^{-1}(y)$, not $X\le g^{-1}(y)$. Continuity of $X$ is why including the endpoint does not change this probability. With atoms, the correct complement uses the left limit of $F_X$.

2. Squaring does not have one global inverse

   For $y\ge0$, $$\{X^2\le y\}=\{-\sqrt y\le X\le\sqrt y\}.$$ Intersect this interval with the input support. Keeping only $+\sqrt y$ loses the negative branch; keeping both branches after one has left the support counts nonexistent outcomes.

Two worked examples

Four graduated exercises

Try each problem before revealing the solution. For every CDF, give its values outside the transformed range as well as inside it.

Before you move on

• Did I intersect each inverse interval with the original support?
• Did I reverse the inequality for a decreasing transformation?
• Did I include every valid inverse branch, and only valid branches?
• Does the CDF have correct boundary values, monotonicity and right-continuity?
• Do the continuous probabilities and any point masses add to one?

Original selected-topic enrichment for probability study alongside MATH 323. This is not an official McGill assessment, a current-term syllabus claim or a complete course review. Check your section’s current outline for the methods expected.

Need help with MATH 323?