I have an Anki deck that I’ve sporadically built over the past two years. It contains cards ranging from Bash commands, to stanzas from Jabberwocky, to questions about using spaced repetition itself. I’ve used it successfully to understand Unix and some CS concepts, but haven’t been intentional about my memory practice habits. I’ve heard the success stories of others, but as with the gym, I am not consistent.

In an effort to change that fact, I’ve been spending time with my now-“angry” Anki deck¹ recently. As a result, I’m finding cards that are so bad that I don’t know how to answer them (how do you create cards for sets vs. enumerations? why did I decide to memorize this vague answer?).

When it comes to my probabilty cards, however, I can feel myself falling into the classic trap. According to Piotr Wozniak, the first rule of spaced repetition is

Do not learn if you do not understand.

While I did spend some time in probability and statistics classes, I don’t understand every probability distribution I’ve added to my deck.

So when this card popped up, I didn’t know where to start.²

Some things I remind myself of whenever I get a card with the word “lognormal” in it:

• When you take the log of a random variable from this distribution, you obtain a random variable that is normally distributed. $$ X \sim \text{Lognormal}(\mu, \sigma^2) \rightarrow \ln(X) = Y \sim N(\mu, \sigma^2) $$
• Lognormal random variables cannot be negative.
• Francis Galton
• …

This blog post is an attempt to get some intuition on the lognormal distribution.

Where does this distribution come from?

Francis Galton studied many naturally occuring phenomena. He got upset when others would measure errors and not take into account that error is not necessarily evenly spaced out. He gave the example of tinting: say you have white scattered on a black background in quantities of $1, 2, 4, 8, …$ These appear to the eye to be equally spaced apart. If someone were to be given a tint of size $4$, they could miss either on the dark side ($2$) or the light side ($8$).

Galton pointed out that the arithmetic mean $\left(\frac{2+8}{2} = 5\right)$ was not the correct way of thinking about this situation: rather, one should use the geometric mean $\left(\sqrt{2*8} = 4\right)$, as the effect of one degree lighter (or darker) was multiplicative, not additive.

What if one took the log of all of these tints? You would end up back with quanitites ($0, 1, 2, 3, …$) where the effect of moving a deviation would be additive again.

Similar to the arithmetic and geometric means in the error above, the normal and lognormal distributions represent the relationship between additive and multiplicative effects. In the normal distribution, the mean $\mu$ is a location parameter. Increasing or decreasing the parameter just shifts the distribution along the $x$-axis. In the lognormal distribution, $e^{\mu}$ is a scaling parameter. Here, increasing $\mu$ changes the shape of the scale of the distribution, not just shifting its center.

What is its density function?

Let $X \sim N(\mu, \sigma^2)$. The probability density function (pdf) of the normal distribution is $$ \large \Phi(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac12\left(\frac{x-\mu}{\sigma}\right)^2\right] $$

Let $Y = e^X$. Then we have $X = \ln(Y)$. It is not immediately intuitive how to derive the lognormal distribution from this information: if you swap $x$ with $ln(y)$, you don’t obtain the correct distribution. Linear transformations are much more straightforward: adding by $C$ or multiplying by $D$ has effects on the distribution that one can derive from their expected value and variance.

The textbook I’m reading doesn’t help with this either. Lawless in “Statistical Models And Methods For Lifetime Data” (p. 24):

and from this p.d.f of $\;T=\exp{Y}$ is easily found to be …

However, if I remembered from probability classes of yore (last year) before looking this up, one can convert between the disribution function (CDF) and density function by integrating, using the desired random variable as the upper bound. The derivation performed here by Gregory Gundersen gives us

$$ \large f_Y(y) = \frac{1}{y\; \sigma \sqrt{2\pi}} \exp\left[ -\frac{1}{2}\left(\frac{\ln y - \mu}{\sigma} \right)^2 \right] $$

How can you get the median from this?

This I also had to look up. I won’t try to derive it now, so I’m left with two options:

• spend the holiday deriving this
• remove it from my Anki deck

Time will tell which I choose.