Computing Conditional Means for Discrete Groups

Welcome. This course begins with the ordinary case in which conditioning really does mean selecting a literal group from a population. It is deliberately familiar territory: if a regressor takes a small set of categories, then $E[Y\mid X=x]$ is just the average outcome among units in category $x$.

That elementary interpretation is the benchmark for the whole course. We will first make it exact for discrete $X$, then ask why the same argument breaks when $X$ is continuous. The later measure-theoretic construction will preserve what matters here: conditional expectation should represent the mean outcome in every group identifiable from $X$.

---

Conditioning means restricting attention to a subpopulation

Let $X$ describe a grouping variable and let $Y$ be a numerical outcome. For example:

• $X$ could be a liquidity regime, classified as low, medium, or high;
• $Y$ could be an order’s implementation shortfall in basis points.

When $X$ is discrete, the event

$$\{X=x\}$$

is a genuine population subgroup. Provided that subgroup has positive probability,

$$P(X=x)>0,$$

we can form a conditional distribution of $Y$ within it:

$$P(Y=y\mid X=x)=\frac{P(Y=y,X=x)}{P(X=x)}.$$

This is simply a renormalisation. The joint probability $P(Y=y,X=x)$ is the share of the whole population that is both in group $x$ and has outcome $y$. Dividing by the share in group $x$ converts those whole-population shares into proportions that sum to one within that group.

The conditional mean is then the ordinary expectation, but calculated using this within-group probability distribution:

$$E[Y\mid X=x]=\sum _{y}yP(Y=y\mid X=x).$$

Equivalently,

$$E[Y\mid X=x]=\sum _{y}y\frac{P(Y=y,X=x)}{P(X=x)}=\frac{{\sum }_{y}yP(Y=y,X=x)}{P(X=x)}.$$

The numerator has an especially useful interpretation:

$$\sum _{y}yP(Y=y,X=x)=E\left[Y{1}_{\{X=x\}}\right].$$

It is the unconditional contribution to expected $Y$ made by the subgroup $X=x$. Dividing by the subgroup’s probability turns that whole-population contribution into its average outcome:

$$\boxed{E[Y\mid X=x]=\frac{E\left[Y{1}_{\{X=x\}}\right]}{P(X=x)}}\text{when }P(X=x)>0.$$

So, in the discrete case, “given $X=x$” has no mystery at all. It means: retain the units for which $X=x$, and average their $Y$-values.

It is worth separating this from a causal claim. The quantity $E[Y\mid X=x]$ describes an association or prediction-relevant group average. It does not, on its own, say what would happen if we intervened to make $X=x$.

---

A discrete trading example

Suppose $X$ is a discrete liquidity classification for an order:

$$X\in \{L,M,H\},$$

where $L$, $M$, and $H$ denote low, medium, and high liquidity stress. Let $Y$ be implementation shortfall, taking the stylised values $0$, $5$, or $10$ basis points. The joint population distribution is:

Liquidity regime $X$	$Y=0$ bps	$Y=5$ bps	$Y=10$ bps	$P(X=x)$
$L$	$0.18$	$0.12$	$0.05$	$0.35$
$M$	$0.10$	$0.20$	$0.10$	$0.40$
$H$	$0.02$	$0.08$	$0.15$	$0.25$

Each interior entry is a joint probability. For instance,

$$P(X=H,Y=10)=\mathrm{0.15.}$$

It says that 15% of all orders are both in the high-stress regime and incur 10 bps of shortfall. It does not say that 15% of high-stress orders incur 10 bps.

To obtain the latter, condition on $X=H$. The high-stress group comprises 25% of all orders, so its conditional distribution is:

$$P(Y=0\mid X=H)=\frac{0.02}{0.25}=0.08,$$ $$P(Y=5\mid X=H)=\frac{0.08}{0.25}=0.32,$$ $$P(Y=10\mid X=H)=\frac{0.15}{0.25}=\mathrm{0.60.}$$

Those probabilities now sum to one because they describe only the high-stress subpopulation.

The conditional mean is the average shortfall within that group:

$$\begin{array}{rl}E[Y\mid X=H]& =0(0.08)+5(0.32)+10(0.60)\\ & =\mathrm{7.6.}\end{array}$$

Thus, the population mean implementation shortfall among high-stress orders is 7.6 bps.

The direct “total divided by group size” calculation gives the same result:

$$\begin{array}{rl}E[Y\mid X=H]& =\frac{0(0.02)+5(0.08)+10(0.15)}{0.25}\\ & =\frac{1.9}{0.25}\\ & =\mathrm{7.6.}\end{array}$$

Imagine a population of 10,000 orders matching these probabilities. There would be 2,500 high-stress orders: 200 with 0 bps shortfall, 800 with 5 bps, and 1,500 with 10 bps. Their total shortfall would be 19,000 bps, and their literal arithmetic average would be:

$$\frac{19,000}{2,500}=7.6\text{ bps}.$$

The probability calculation is simply the population-level version of this group average.

---

From several group means to a conditional-mean function

The previous calculation gives one number, $E[Y\mid X=H]$. Repeating it for each possible value of $X$ gives:

Group $x$	$P(X=x)$	$E[Y\mid X=x]$
$L$	$0.35$	$3.14$ bps
$M$	$0.40$	$5.00$ bps
$H$	$0.25$	$7.60$ bps

For low stress, for example,

$$E[Y\mid X=L]=\frac{0(0.18)+5(0.12)+10(0.05)}{0.35}=\frac{22}{7}.$$

The mapping

$$m(x)=E[Y\mid X=x]$$

is the conditional mean function or, in regression language, the population regression function. In this discrete example it is simply a lookup table: enter the observed liquidity regime and retrieve that group’s population mean shortfall.

There are three closely related expressions worth keeping distinct:

$$E[Y]$$

is one unconditional population mean.

$$E[Y\mid X=H]$$

is one particular group mean.

$$E[Y\mid X]$$

is shorthand for the random quantity that takes the value appropriate to the realised group. In this example,

$$E[Y\mid X]=\{\begin{array}{ll}3.14,& X=L,\\ 5.00,& X=M,\\ 7.60,& X=H.\end{array}$$

This notation is natural: before you observe $X$, you do not yet know which conditional mean applies. Once $X$ is observed, the relevant group mean is determined.

The unconditional mean is recovered by averaging the group means using the group probabilities:

$$\begin{array}{rl}E[Y]& =\sum _{x}E[Y\mid X=x]P(X=x)\\ & =(3.14)(0.35)+(5)(0.40)+(7.6)(0.25)\\ & =5\text{ bps}.\end{array}$$

This is the law of total expectation in the discrete setting. It says that the overall average is a weighted average of subgroup averages. The weights are not arbitrary: large groups contribute more to the population average than small groups.

The decomposition also clarifies what conditioning adds. Without liquidity information, the best single summary of the population is 5 bps. With the discrete $X$-information, the relevant average varies from roughly 3.14 bps to 7.6 bps. Conditioning replaces one pooled average with averages tailored to identifiable groups.

---

The sample analogue, and the denominator that matters

In a dataset of observations $(X_i,Y_i)$, the empirical counterpart of the population conditional mean is exactly the group average:

$$\hat{m}(x)=\frac{{\sum }_{i=1}^n{Y}_i{1}_{\{X_i=x\}}}{{\sum }_{i=1}^n{1}_{\{X_i=x\}}},$$

provided at least one observation belongs to group $x$.

The denominator is the number of observations in the group. The numerator is their total $Y$-value. Hence the formula is simply “sum within group divided by count within group.”

The population formula has the same structure:

$$E[Y\mid X=x]=\frac{E[Y{1}_{\{X=x\}}]}{P(X=x)}.$$

Here, $P(X=x)$ is the population analogue of the sample group proportion. This is why discrete conditioning is so intuitively secure: both in data and in the population model, there is a nonempty group over which an average can be taken.

There is one condition that should not be treated as a technical footnote:

$$P(X=x)>0.$$

If a supposed category has zero population probability, there is no population subgroup to average over, and the ratio definition is undefined. In a discrete model, that usually just means $x$ is outside the support of $X$, or is an impossible state.

The next lesson considers a more consequential case: $X$ is continuous, so every exact numerical value has probability zero even though such values can be observed. The familiar group-average formula then presents a $0/0$ problem. The rest of the course is about retaining the substantive meaning of conditional averages without pretending that this ratio still works point by point.

---

Key takeaways

For a discrete conditioning variable $X$, $E[Y\mid X=x]$ is the population mean of $Y$ within the genuine subgroup $\{X=x\}$:

$$E[Y\mid X=x]=\sum _{y}yP(Y=y\mid X=x).$$

It can also be written as a ratio of the subgroup’s total expected outcome to its population share:

$$E[Y\mid X=x]=\frac{E[Y{1}_{\{X=x\}}]}{P(X=x)}.$$

As $x$ varies, these group means form the conditional-mean function $m(x)$. Their probability-weighted average recovers $E[Y]$.

Next, we will isolate the exact point at which this elementary reasoning fails for a continuous regressor: the event $\{X=x\}$ no longer has positive probability, so the literal “select the group and average it” definition cannot be used unchanged.