# My Favorite One-Liners: Part 43

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. q Q

Years ago, my first class of students decided to call me “Dr. Q” instead of “Dr. Quintanilla,” and the name has stuck ever since. And I’ll occasionally use this to my advantage when choosing names of variables. For example, here’s a typical proof by induction involving divisibility.

Theorem: If $n \ge 1$ is a positive integer, then $5^n - 1$ is a multiple of 4.

Proof. By induction on $n$.

$n = 1$: $5^1 - 1 = 4$, which is clearly a multiple of 4.

$n$: Assume that $5^n - 1$ is a multiple of 4.

At this point in the calculation, I ask how I can write this statement as an equation. Eventually, somebody will volunteer that if $5^n-1$ is a multiple of 4, then $5^n-1$ is equal to 4 times something. At which point, I’ll volunteer:

Yes, so let’s name that something with a variable. Naturally, we should choose something important, something regal, something majestic… so let’s choose the letter $q$. (Groans and laughter.) It’s good to be the king.

So the proof continues:

$n$: Assume that $5^n - 1 = 4q$, where $q$ is an integer.

$n+1$. We wish to show that $5^{n+1} - 1$ is also a multiple of 4.

At this point, I’ll ask my class how we should write this. Naturally, I give them no choice in the matter:

We wish to show that $5^{n+1} - 1 = 4Q$, where $Q$ is some (possibly different) integer.

Then we continue the proof:

$5^{n+1} - 1 = 5^n 5^1 - 1$

$= 5 \times 5^n - 1$

$= 5 \times (4q + 1) - 1$ by the induction hypothesis

$= 20q + 5 - 1$

$= 20q + 4$

$= 4(5q + 1)$.

So if we let $Q = 5q +1$, then $5^{n+1} - 1 = 4Q$, where $Q$ is an integer because $q$ is also an integer.

QED

On the flip side of braggadocio, the formula for the binomial distribution is

$P(X = k) = \displaystyle {n \choose k} p^k q^{n-k}$,

where $X$ is the number of successes in $n$ independent and identically distributed trials, where $p$ represents the probability of success on any one trial, and (to my shame) $q$ is the probability of failure.

# My Favorite One-Liners: Part 40

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In some classes, the Greek letter $\phi$ or $\Phi$ naturally appears. Sometimes, it’s an angle in a triangle or a displacement when graphing a sinusoidal function. Other times, it represents the cumulative distribution function of a standard normal distribution.

Which begs the question, how should a student pronounce this symbol?

I tell my students that this is the Greek letter “phi,” pronounced “fee”. However, other mathematicians may pronounce it as “fie,” rhyming with “high”. Continuing,

Other mathematicians pronounce it as “foe.” Others, as “fum.”

# My Favorite One-Liners: Part 33

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Perhaps one of the more difficult things that I try to instill in my students is numeracy, or a sense of feeling if an answer to a calculation is plausible. As a initial step toward this goal, I’ll try to teach my students some basic pointers about whether an answer is even possible.

For example, when calculating a standard deviation, students have to compute $E(X)$ and $E(X^2)$:

$E(X) = \sum x p(x) \qquad \hbox{or} \qquad E(X) = \int_a^b x f(x) \, dx$

$E(X^2) = \sum x^2 p(x) \qquad \hbox{or} \qquad E(X^2) = \int_a^b x^2 f(x) \, dx$

After these are computed — which could take some time — the variance is then calculated:

$\hbox{Var}(X) = E(X^2) - [E(X)]^2$.

Finally, the standard deviation is found by taking the square root of the variance.

So, I’ll ask my students, what do you do if you calculate the variance and it’s negative, so that it’s impossible to take the square root? After a minute to students hemming and hawing, I’ll tell them emphatically what they should do:

It’s wrong… do it again.

The same principle applies when computing probabilities, which always have to be between 0 and 1. So, if ever a student computes a probability that’s either negative or else greater than 1, they can be assured that the answer is wrong and that there’s a mistake someplace in their computation that needs to be found.

# My Favorite One-Liners: Part 31

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s the closing example that I’ll use when presenting the binomial and hypergeometric distributions to my probability/statistics students.

A lonely bachelor decides to play the field, deciding that a lifetime of watching “Leave It To Beaver” reruns doesn’t sound all that pleasant. On 250 consecutive days, he calls a different woman for a date. Unfortunately, through the school of hard knocks, he knows that the probability that a given woman will accept his gracious invitation is only 1%. What is the chance that he will land at least three dates?

You can probably imagine the stretch I was enduring when I first developed this example many years ago. Nevertheless, I make a point to add the following disclaimer before we start finding the solution, which always gets a laugh:

The events of this exercise are purely fictitious. Any resemblance to any actual persons — living, or dead, or currently speaking — is purely coincidental.

# My Favorite One-Liners: Part 30

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s quip is a follow-up to yesterday’s post and is one that I’ll use when I need my students to remember something that I taught them earlier in the semester — perhaps even the previous day.

For example, in my applied statistics class, one day I’ll show students how to compute the expected value and the standard deviation of a random variable:

$E(X) = \sum x \cdot P(X=x)$

$E(X^2) = \sum x^2 \cdot P(X=x)$

$\hbox{SD}(X) = \sqrt{ E(X^2) - [E(X)]^2 }$

Then, the next time I meet them, I start working on a seemingly new topic, the derivation of the binomial distribution:

$P(X = k) = \displaystyle {n \choose k} p^k q^{n-k}$.

This derivation takes some time because I want my students to understand not only how to use the formula but also where the formula comes from. Eventually, I’ll work out that if $n = 3$ and $p = 0.2$,

$P(X = 0) = 0.512$

$P(X = 1) = 0.384$

$P(X = 2) = 0.096$

$P(X = 3) = 0.008$

Then, I announce to my class, I next want to compute $E(X)$ and $\hbox{SD}(X)$. We had just done this the previous class period; however, I know full well that they haven’t yet committed those formulas to memory. So here’s the one-liner that I use: “If you had a good professor, you’d remember how to do this.”

Eventually, when the awkward silence has lasted long enough because no one can remember the formula (without looking back at the previous day’s notes), I plunge an imaginary knife into my heart and turn the imaginary dagger, getting the point across: You really need to remember this stuff.

# My Favorite One-Liners: Part 29

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s quip is one that I’ll use when I need my students to remember something from a previous course — especially when it’s a difficult concept from a previous course — that somebody else taught them in a previous semester.

For example, in my probability class, I’ll introduce the Poisson distribution

$P(X = k) = e^{-\mu} \displaystyle \frac{\mu^k}{k!}$,

where $\mu > 0$ and the permissible values of $k$ are non-negative integers.

In particular, since these are probabilities and one and only one of these values can be taken, this means that

$\displaystyle \sum_{k=0}^\infty e^{-\mu} \frac{\mu^k}{k!} = 1$.

At this point, I want students to remember that they’ve actually seen this before, so I replace $\mu$ by $x$ and then multiply both sides by $e^x$:

$\displaystyle \sum_{k=0}^\infty \frac{x^k}{k!} = e^x$.

Of course, this is the Taylor series expansion for $e^x$. However, my experience is that most students have decidedly mixed feelings about Taylor series; often, it’s the last thing that they learn in Calculus II, which means it’s the first thing that they forget when the semester is over. Also, most students have a really hard time with Taylor series when they first learn about them.

So here’s my one-liner that I’ll say at this point: “Does this bring back any bad memories for anyone? Perhaps like an old Spice Girls song?” And this never fails to get an understanding laugh before I remind them about Taylor series.

# My Favorite One-Liners: Part 26

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that could appear early in a probability class:

Let $P(A) = 0.2$, $P(B) = 0.4$, and $P(A \cup B) = 0.5$. Find $P(A \mid B)$.

The standard technique for solving this problem involves first finding $P(A \cap B)$ using the Addition Rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$0.5 = 0.2 + 0.4 - P(A \cap B)$

$P(A \cap B) = 0.1$

From here, the Multiplication Rule can be used (or, equivalently, the definition of a conditional probability):

$P(B \cap A) = P(B) \cdot P(A \mid B)$

$0.1 = 0.4 P(A \mid B)$

$0.25 = P(A \mid B)$

So far, so good.

Now let me add a small twist to the original problem that creates a small difficulty when solving:

Let $P(A) = 0.2$, $P(B) = 0.4$, and $P(A \cup B) = 0.5$. Find $P(A \cap B \mid A \cup B)$.

Proceeding as before, we obtain

$P( [A \cap B] \cup [A \cup B] ) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)$

The value of $P(A \cup B)$ is obvious. But how do we evaluate the left side?

If I’m teaching an advanced probability class, I might expect them to use DeMorgan’s Laws. However, it’s a whole lot easier to reason out the left hand side: I’m looking for the probability that both $A$ and $B$ happen or else at least one of $A$ and $B$ happen. Well, that’s clearly redundant: if both $A$ and $B$ happen, then certainly at least one of $A$ and $B$ happen.

Here’s my one-liner, which I say, if possible, using only one breath of air:

Clearly, this is redundant. It’s like saying Dr. Q is my professor and he’s a total stud. It’s redundant. It’s obvious. There’s no need to actually say it.

After the laughter settles from this bit of braggadocio, the $A \cup B$ can be safely dropped from the left side:

$P( A \cap B) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)$

$0.1 = 0.5 \cdot P(A \cap B \mid A \cup B)$

$0.2 = P(A \cap B \mid A \cup B)$

However, I need to emphasize that dropping the term on the left side is a special feature of this particular problem since one set was a subset of the other, and that students shouldn’t expect to always be able to do this when computing conditional probabilities.

# My Favorite One-Liners: Part 24

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that could appear in my class in probability or statistics:

Let $f(x) = 3x^2$ be a probability density function for $0 \le x \le 1$. Find $F(x) = P(X \le x)$, the cumulative distribution function of $X$.

A student’s first reaction might be to set up the integral as

$\displaystyle \int_0^x 3x^2 \, dx$

The problem with this set-up, of course, is that the letter $x$ has already been reserved as the right endpoint for this definite integral. Therefore, inside the integral, we should choose any other letter — just not $x$ — as the dummy variable.

Which sets up my one-liner: “In the words of the great philosopher Jean-Luc Picard: Plenty of letters left in the alphabet.”

We then write the integral as something like

$\displaystyle \int_0^x 3t^2 \, dt$

and then get on with the business of finding $F(x)$.

# My Favorite One-Liners: Part 13

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a story that I’ll tell my students when, for the first time in a semester, I’m about to use a previous theorem to make a major step in proving a theorem. For example, I may have just finished the proof of

$\hbox{Var}(X+Y) = \hbox{Var}(X) + \hbox{Var}(Y)$,

where $X$ and $Y$ are independent random variables, and I’m about to prove that

$\hbox{Var}(X-Y) = \hbox{Var}(X) + \hbox{Var}(Y)$.

While this can be done by starting from scratch and using the definition of variance, the easiest thing to do is to write

$\hbox{Var}(X-Y) = \hbox{Var}(X+[-Y]) = \hbox{Var}(X) + \hbox{Var}(-Y)$,

thus using the result of the first theorem to prove the next theorem.

And so I have a little story that I tell students about this principle. I think I was 13 when I first heard this one, and obviously it’s stuck with me over the years.

At MIT, there’s a two-part entrance exam to determine who will be the engineers and who will be the mathematicians. For the first part of the exam, students are led one at a time into a kitchen. There’s an empty pot on the floor, a sink, and a stove. The assignment is to boil water. Everyone does exactly the same thing: they fill the pot with water, place it on the stove, and then turn the stove on. Everyone passes.

For the second part of the exam, students are led one at a time again into the kitchen. This time, there’s a pot full of water sitting on the stove. The assignment, once again, is to boil water. Nearly everyone simply turns on the stove. These students are led off to become engineers. The mathematicians are ones who take the pot off the stove, dump the water into the sink, and place the empty pot on the floor… thereby reducing to the original problem, which had already been solved.

# My Favorite One-Liners: Part 7

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

One of my favorite one-liners is simply stated: “Mathematicians are lazy.” I’ll use this whenever I introduce my students a new piece of mathematical notation or lingo.

For example, in probability, a common notion is a sequence of independent and identically distributed random variables (say, rolling a die repeatedly). However, mathematicians will typically write “i.i.d.” instead of “independent and identically distributed.” Why? That’s when I break out the mantra: “Mathematicians are lazy.” It’s my quick way of saying, “Hey, this is new notation that you’re about to learn, but the whole point of new notation is to make writing mathematical ideas a little quicker.”

Mathematical notation like $\displaystyle \int_a^b f(x) \, dx$ can appear very intimidating when students first encounter them. Hopefully repeating this mantra a few dozen times each semester makes the introduction of new notation a little more palatable for my students.