The Central Idea
Let \(X\) be a random variable and \(B\) an event with
\[ P(B)>0. \]The conditional expectation
\[ \boxed{E[X\mid B]} \]is the expected value of \(X\) when we know that \(B\) occurred.
The probabilities used in the average are conditional probabilities rather than the original unconditional probabilities.
Ordinary Expectation
For a discrete random variable \(X\) taking values
\[ x_1,x_2,\ldots,x_n, \]the ordinary expected value is
\[ \boxed{ E[X] = \sum_{i=1}^{n}x_iP(X=x_i) }. \]Each possible value is weighted by its unconditional probability.
Conditional Expectation for a Discrete Random Variable
Given an event \(B\) with positive probability,
\[ \boxed{ E[X\mid B] = \sum_i x_iP(X=x_i\mid B) }. \]The structure is the same as ordinary expectation.
The only change is that the weights are conditional probabilities.
Conditioning Changes the Distribution
Before conditioning, the random variable has probabilities
\[ P(X=x_i). \]After learning \(B\), the relevant probabilities become
\[ P(X=x_i\mid B). \]Conditional expectation averages the same numerical values using this updated probability distribution.
A First Example
Roll a fair six-sided die and let \(X\) be the number rolled.
The ordinary expectation is
\[ E[X] = \frac{1+2+3+4+5+6}{6} = \boxed{3.5}. \]Now suppose we learn that the roll was even.
Condition on an Even Result
Let
\[ B=\{2,4,6\}. \]Given \(B\), the three remaining outcomes are equally likely.
Therefore,
\[ E[X\mid B] = 2\left(\frac13\right) + 4\left(\frac13\right) + 6\left(\frac13\right). \]Thus,
\[ \boxed{ E[X\mid B]=4 }. \]Knowing that the result is even raises the expected value from \(3.5\) to \(4\).
Conditioning Can Raise or Lower an Expectation
If we instead condition on
\[ C=\{1,2,3\}, \]then
\[ E[X\mid C] = \frac{1+2+3}{3} = \boxed{2}. \]The condition selects lower-valued outcomes, so the conditional expectation is below the unconditional expectation.
Conditional Expectation Is a Weighted Average
The value
\[ E[X\mid B] \]is a weighted average of the possible values of \(X\) inside the conditioned event \(B\).
Therefore, it must lie between the smallest and largest values of \(X\) that remain possible under \(B\).
Example with Unequal Conditional Probabilities
Suppose a random variable \(X\) takes values
\[ 10,\quad20,\quad30 \]under a condition \(B\), with
\[ P(X=10\mid B)=0.20, \] \[ P(X=20\mid B)=0.50, \] \[ P(X=30\mid B)=0.30. \]Then
\[ E[X\mid B] = 10(0.20)+20(0.50)+30(0.30). \]Therefore,
\[ \boxed{ E[X\mid B]=21 }. \]Conditional Probabilities Must Sum to One
Within the condition \(B\),
\[ \sum_i P(X=x_i\mid B)=1. \]This is why conditional expectation is a genuine weighted average.
Compute Conditional Probabilities First
Sometimes the problem gives joint probabilities rather than conditional probabilities.
Then use
\[ P(X=x_i\mid B) = \frac{P(X=x_i\cap B)}{P(B)}. \]These conditional probabilities can then be inserted into the expectation formula.
Equivalent Joint-Probability Formula
Substitute the conditional probability formula:
\[ E[X\mid B] = \sum_i x_i \frac{P(X=x_i\cap B)}{P(B)}. \]Since \(P(B)\) is constant with respect to the sum,
\[ \boxed{ E[X\mid B] = \frac{\sum_i x_iP(X=x_i\cap B)}{P(B)} }. \]Example from Joint Probabilities
Suppose \(X\) takes values \(1,2,3\), and for an event \(B\),
\[ P(X=1\cap B)=0.10, \] \[ P(X=2\cap B)=0.20, \] \[ P(X=3\cap B)=0.30. \]Then
\[ P(B)=0.10+0.20+0.30=0.60. \]Continue the Joint-Probability Example
Using the joint-probability formula,
\[ E[X\mid B] = \frac{ 1(0.10)+2(0.20)+3(0.30) }{0.60}. \]The numerator is
\[ 0.10+0.40+0.90=1.40. \]Therefore,
\[ \boxed{ E[X\mid B]=\frac{1.40}{0.60}=\frac73 }. \]Conditional Expectation from a Table
Suppose \(X\) is the number of purchases made by a customer in a month, and customers are classified as premium or standard.
| Purchases \(X\) | Premium | Standard |
|---|---|---|
| 0 | 10 | 30 |
| 1 | 20 | 40 |
| 2 | 30 | 20 |
| 3 | 40 | 10 |
Find the Conditional Distribution for Premium Customers
The premium column contains
\[ 10+20+30+40=100 \]customers.
Therefore,
\[ P(X=0\mid P)=0.10, \] \[ P(X=1\mid P)=0.20, \] \[ P(X=2\mid P)=0.30, \] \[ P(X=3\mid P)=0.40. \]Find the Premium Conditional Expectation
Compute
\[ E[X\mid P] = 0(0.10)+1(0.20)+2(0.30)+3(0.40). \]Therefore,
\[ \boxed{ E[X\mid P]=2 }. \]Premium customers average \(2\) purchases per month in this table.
Find the Standard Conditional Expectation
The standard column also contains \(100\) customers.
Thus,
\[ E[X\mid S] = 0(0.30)+1(0.40)+2(0.20)+3(0.10). \]Therefore,
\[ \boxed{ E[X\mid S]=1.1 }. \]The expected number of purchases differs across the two customer groups.
Conditional Expectation by Group
In applied work, conditional expectations often appear as subgroup averages.
Examples include:
- expected revenue given customer type;
- expected loss given risk class;
- expected waiting time given service channel;
- expected score given completion status;
- expected lifetime given component category.
Conditional Expectation and Partitions
Suppose
\[ B_1,B_2,\ldots,B_n \]form a partition of the sample space.
Then each conditional expectation
\[ E[X\mid B_i] \]describes the expected value within one partition group.
The Law of Total Expectation
If \(B_1,\ldots,B_n\) form a partition, then
\[ \boxed{ E[X] = \sum_{i=1}^{n} P(B_i)E[X\mid B_i] }. \]This is the law of total expectation.
The overall expectation is a weighted average of the conditional expectations.
The Two-Group Version
For an event \(B\) and its complement \(B^c\),
\[ \boxed{ E[X] = P(B)E[X\mid B] + P(B^c)E[X\mid B^c] }. \]This parallels the two-case law of total probability.
Example: Customer Segments
Suppose \(40\%\) of customers are premium and \(60\%\) are standard:
\[ P(P)=0.40, \qquad P(S)=0.60. \]Suppose expected monthly purchases are
\[ E[X\mid P]=5 \]and
\[ E[X\mid S]=2. \]Find the Overall Expected Purchases
Apply the law of total expectation:
\[ E[X] = 0.40(5)+0.60(2). \]Therefore,
\[ E[X] = 2+1.2 = \boxed{3.2}. \]The overall expected number of monthly purchases is \(3.2\).
Conditional Expectations Act Like Group Means
In the previous example,
\[ E[X\mid P]=5 \]is the expected value inside the premium group, while
\[ E[X\mid S]=2 \]is the expected value inside the standard group.
The overall mean is obtained by weighting these group means by group size.
Example: Manufacturing Output
Suppose \(70\%\) of units come from Line 1 and \(30\%\) from Line 2.
Let \(X\) be repair cost per unit.
Suppose
\[ E[X\mid L_1]=\$4 \]and
\[ E[X\mid L_2]=\$10. \]Then
\[ E[X] = 0.70(4)+0.30(10) = \boxed{\$5.80}. \]Example: Insurance Loss
Suppose an insurer divides policies into low-risk and high-risk classes.
Let \(X\) be annual claim cost.
Suppose
\[ P(L)=0.80, \qquad P(H)=0.20, \] \[ E[X\mid L]=\$500, \] \[ E[X\mid H]=\$2{,}000. \]Then
\[ E[X] = 0.80(500)+0.20(2000) = \boxed{\$800}. \]Example: Conditional Profit
Suppose a project earns
\[ \$100{,}000 \]if market demand is high and loses
\[ \$20{,}000 \]if demand is low.
After observing new information \(B\), suppose
\[ P(\text{high}\mid B)=0.70. \]Then
\[ P(\text{low}\mid B)=0.30. \]Find the Conditional Expected Profit
Let \(X\) be project profit.
Then
\[ E[X\mid B] = 100000(0.70) + (-20000)(0.30). \]Therefore,
\[ \boxed{ E[X\mid B]=\$64{,}000 }. \]The expected value has been updated using the conditional probabilities implied by the new information.
Conditional Expectation Can Use Bayes' Theorem
Sometimes the conditional probabilities needed for
\[ E[X\mid B] \]are not given directly.
Bayes' theorem may first be used to compute
\[ P(X=x_i\mid B), \]after which those probabilities are used as expectation weights.
Example: Hidden Source and Observed Evidence
Suppose a unit comes from Source 1 with probability
\[ 0.60 \]and Source 2 with probability
\[ 0.40. \]Let repair cost \(X\) equal
\[ \$100 \]for Source 1 and
\[ \$300 \]for Source 2.
Suppose a warning signal \(W\) occurs with probabilities
\[ P(W\mid S_1)=0.20, \qquad P(W\mid S_2)=0.60. \]Find the Probability of the Warning
By total probability,
\[ P(W) = 0.60(0.20)+0.40(0.60). \]Therefore,
\[ P(W) = 0.12+0.24 = \boxed{0.36}. \]Update the Source Probabilities
By Bayes' theorem,
\[ P(S_1\mid W) = \frac{0.60(0.20)}{0.36} = \frac13, \]and
\[ P(S_2\mid W) = \frac{0.40(0.60)}{0.36} = \frac23. \]Find the Conditional Expected Repair Cost
Given the warning,
\[ E[X\mid W] = 100\left(\frac13\right) + 300\left(\frac23\right). \]Therefore,
\[ \boxed{ E[X\mid W]=\$233.33\text{ approximately} }. \]The observed warning changes the expected repair cost because it changes the source probabilities.
Indicator Random Variables
For an event \(A\), define its indicator random variable
\[ I_A = \begin{cases} 1,&A\text{ occurs},\\ 0,&A\text{ does not occur}. \end{cases} \]Then
\[ \boxed{ E[I_A]=P(A) }. \]Conditional Expectation of an Indicator
Given an event \(B\) with positive probability,
\[ \boxed{ E[I_A\mid B]=P(A\mid B) }. \]This shows that conditional probability is a special case of conditional expectation.
Why the Indicator Identity Works
Under the condition \(B\), the indicator \(I_A\) equals \(1\) when \(A\) occurs and \(0\) otherwise.
Therefore,
\[ E[I_A\mid B] = 1\cdot P(A\mid B) + 0\cdot P(A^c\mid B). \]Thus,
\[ E[I_A\mid B] = P(A\mid B). \]Linearity of Conditional Expectation
For constants \(a\) and \(b\),
\[ \boxed{ E[aX+bY\mid B] = aE[X\mid B]+bE[Y\mid B] }. \]Conditional expectation preserves the familiar linearity property of ordinary expectation.
Adding a Constant
For a constant \(c\),
\[ E[X+c\mid B] = E[X\mid B]+c. \]A constant remains unchanged by averaging.
Multiplying by a Constant
For a constant \(c\),
\[ E[cX\mid B] = cE[X\mid B]. \]The scale factor can be pulled outside the conditional expectation.
Conditional Expectation of a Constant
If \(c\) is constant, then
\[ \boxed{ E[c\mid B]=c }. \]Conditioning does not change a quantity that is already fixed.
Independence and Conditional Expectation
If the distribution of \(X\) is unaffected by whether \(B\) occurs, then
\[ P(X=x_i\mid B)=P(X=x_i) \]for all relevant values.
In that case,
\[ \boxed{ E[X\mid B]=E[X] }. \]The condition provides no information that changes the expected value.
Example: Independent Die and Coin
Roll a fair die and independently toss a fair coin.
Let \(X\) be the die result and let \(H\) be the event that the coin is heads.
Since the coin outcome does not affect the die distribution,
\[ E[X\mid H]=E[X]=\boxed{3.5}. \]Dependence Can Change Conditional Expectation
If the condition changes the probability distribution of \(X\), then
\[ E[X\mid B] \]can differ from
\[ E[X]. \]Conditional expectation therefore provides a numerical summary of how information changes the distribution's center.
Example: Expected Waiting Time by Channel
Suppose \(X\) is customer waiting time in minutes.
For phone support,
\[ E[X\mid P]=12. \]For chat support,
\[ E[X\mid C]=4. \]The conditional expectations quantify the different average waiting times within the two channels.
Example: Weighted Overall Waiting Time
Suppose \(30\%\) of customers use phone support and \(70\%\) use chat.
Then
\[ E[X] = 0.30(12)+0.70(4). \]Therefore,
\[ \boxed{ E[X]=6.4 }. \]The overall expected waiting time is \(6.4\) minutes.
Conditional Expectation Can Be a Random Variable
In more advanced settings, conditional expectation may be conditioned on a random variable rather than on a single event.
The notation
\[ E[X\mid Y] \]represents an expectation that depends on the observed value of \(Y\).
As \(Y\) varies, the conditional expectation can vary, so
\[ E[X\mid Y] \]is itself a random variable.
Conditioning on a Discrete Random Variable
If \(Y\) is discrete, then for each value \(y\),
\[ \boxed{ E[X\mid Y=y] = \sum_x xP(X=x\mid Y=y) }. \]This gives one conditional expected value for each possible value of \(Y\).
Example: Conditional Expectation as a Function
Suppose
\[ E[X\mid Y=0]=2 \]and
\[ E[X\mid Y=1]=5. \]Then the random quantity
\[ E[X\mid Y] \]equals \(2\) when \(Y=0\) and \(5\) when \(Y=1\).
The Tower Property
A central relationship is
\[ \boxed{ E[E[X\mid Y]]=E[X] }. \]This is often called the law of iterated expectation or the tower property.
It generalizes the law of total expectation.
Why the Tower Property Makes Sense
First compute the expected value of \(X\) within each value or information state of \(Y\).
Then average those conditional expectations over the distribution of \(Y\).
The result is the original unconditional expectation of \(X\).
Example: Tower Property with Two Groups
Suppose
\[ P(Y=0)=0.70, \qquad P(Y=1)=0.30, \]with
\[ E[X\mid Y=0]=2 \]and
\[ E[X\mid Y=1]=6. \]Then
\[ E[X] = 0.70(2)+0.30(6) = \boxed{3.2}. \]Conditional Expectation as Information Compression
A full conditional distribution may contain many probabilities.
The conditional expectation compresses that distribution into one number: its mean.
This summary is useful, but it does not preserve every feature of the conditional distribution.
Equal Conditional Expectations Do Not Imply Equal Distributions
Two groups can have the same conditional expectation while having very different probability distributions.
Therefore,
\[ E[X\mid B_1]=E[X\mid B_2] \]does not imply that
\[ P(X=x\mid B_1)=P(X=x\mid B_2) \]for every \(x\).
Example: Same Mean, Different Conditional Distributions
Group 1 gives
\[ X=0\text{ or }10 \]with equal probability.
Then
\[ E[X\mid B_1]=5. \]Group 2 always gives
\[ X=5. \]Then
\[ E[X\mid B_2]=5. \]The conditional means match even though the distributions are very different.
Expected Value Is Not a Guaranteed Outcome
A conditional expectation is an average under a conditional probability model.
It need not be one of the possible observed values.
For example, a fair die conditioned on being greater than \(3\) has
\[ E[X\mid X>3] = \frac{4+5+6}{3} = \boxed{5}. \]In other examples, the conditional expectation may be a non-integer value even when all outcomes are integers.
Example: Non-Integer Conditional Expectation
Suppose, under condition \(B\),
\[ P(X=1\mid B)=0.25 \]and
\[ P(X=2\mid B)=0.75. \]Then
\[ E[X\mid B] = 1(0.25)+2(0.75) = \boxed{1.75}. \]The value \(1.75\) is not itself a possible outcome of \(X\).
Conditional Expectation in Decision Models
Conditional expectation often appears in decision analysis because new information changes the probabilities of future payoffs.
A decision maker may compare
\[ E[X] \]with
\[ E[X\mid B] \]to understand how information changes the expected outcome.
Conditional Expectation Does Not Automatically Determine the Best Decision
Expected value is one quantitative criterion.
A real decision may also depend on:
- risk tolerance;
- constraints;
- tail outcomes;
- liquidity;
- timing;
- policy or operational requirements.
Conditional expectation describes an updated average, not a complete decision rule.
Common Error: Using Unconditional Probabilities After Conditioning
Once \(B\) is known, the expectation must use
\[ P(X=x_i\mid B), \]not the original
\[ P(X=x_i). \]Conditioning changes the weights.
Common Error: Forgetting to Renormalize
If some outcomes are removed by the condition, the remaining probabilities must sum to \(1\).
It is not enough simply to delete excluded outcomes while leaving the old probabilities unchanged.
Common Error: Averaging Conditional Means Without Weights
Suppose
\[ E[X\mid B_1]=2 \]and
\[ E[X\mid B_2]=8. \]The overall expectation is not automatically
\[ \frac{2+8}{2}=5. \]The group probabilities must be used:
\[ E[X] = P(B_1)E[X\mid B_1] + P(B_2)E[X\mid B_2]. \]Common Error: Confusing Conditional Expectation with Conditional Probability
Conditional probability,
\[ P(A\mid B), \]is a probability between \(0\) and \(1\).
Conditional expectation,
\[ E[X\mid B], \]is an average of the values of \(X\) and may lie outside the interval \([0,1]\).
Common Error: Treating the Conditional Mean as Certain
The value
\[ E[X\mid B] \]summarizes the conditional distribution.
It does not say that \(X\) must equal that value whenever \(B\) occurs.
Common Error: Assuming Equal Means Imply Independence
If
\[ E[X\mid B]=E[X], \]that equality alone does not necessarily imply that \(X\) and \(B\) are independent.
Independence requires the full conditional distribution to remain unchanged, not merely its mean.
Common Error: Ignoring the Condition Probability
The event-based conditional expectation
\[ E[X\mid B] \]requires
\[ P(B)>0. \]The elementary event-based definition is not available when the conditioning event has probability zero.
A Reliable Procedure
- Identify the random variable \(X\).
- Identify the condition \(B\).
- List the values of \(X\) still possible under \(B\).
- Compute the conditional probabilities \(P(X=x_i\mid B)\).
- Check that those conditional probabilities sum to \(1\).
- Multiply each possible value by its conditional probability.
- Add the products.
- Interpret the result as an average under the conditioned information.
A Reliable Procedure for Total Expectation
- Identify a partition \(B_1,\ldots,B_n\).
- Find each group probability \(P(B_i)\).
- Find each conditional mean \(E[X\mid B_i]\).
- Multiply \(P(B_i)E[X\mid B_i]\) for every group.
- Add the weighted conditional means.
- Check that the group probabilities sum to \(1\).
- Interpret the result as the overall expectation.
Example: Full Conditional-Distribution Workflow
Suppose a random variable \(X\) has the following conditional distribution given \(B\):
| \(x\) | \(P(X=x\mid B)\) |
|---|---|
| 0 | 0.10 |
| 5 | 0.30 |
| 10 | 0.60 |
Then
\[ E[X\mid B] = 0(0.10)+5(0.30)+10(0.60). \]Therefore,
\[ \boxed{ E[X\mid B]=7.5 }. \]Example: Full Total-Expectation Workflow
Suppose customers are divided into three segments:
\[ P(S_1)=0.50, \qquad P(S_2)=0.30, \qquad P(S_3)=0.20. \]Expected annual revenue per customer is
\[ E[X\mid S_1]=100, \] \[ E[X\mid S_2]=200, \] \[ E[X\mid S_3]=500. \]Then
\[ E[X] = 0.50(100)+0.30(200)+0.20(500). \]Thus,
\[ E[X] = 50+60+100 = \boxed{210}. \]Practice 1: Conditioned Die
A fair die is rolled. Let \(X\) be the result. Given that \(X>3\), find
\[ E[X\mid X>3]. \]Practice 2: Conditional Distribution
Given \(B\), suppose
\[ P(X=1\mid B)=0.20, \] \[ P(X=2\mid B)=0.50, \] \[ P(X=5\mid B)=0.30. \]Find \(E[X\mid B]\).
Practice 3: Conditional Expectation from Counts
Among \(100\) premium customers:
- \(20\) make \(0\) purchases;
- \(30\) make \(1\) purchase;
- \(50\) make \(2\) purchases.
Let \(X\) be number of purchases. Find \(E[X\mid P]\).
Practice 4: Total Expectation
Suppose
\[ P(B)=0.40, \qquad P(B^c)=0.60, \] \[ E[X\mid B]=8, \]and
\[ E[X\mid B^c]=3. \]Find \(E[X]\).
Practice 5: Three Groups
Suppose
\[ P(B_1)=0.50, \quad P(B_2)=0.30, \quad P(B_3)=0.20, \]and
\[ E[X\mid B_1]=2, \quad E[X\mid B_2]=5, \quad E[X\mid B_3]=10. \]Find \(E[X]\).
Practice 6: Indicator Variable
Let \(I_A\) be the indicator of event \(A\). If
\[ P(A\mid B)=0.65, \]find
\[ E[I_A\mid B]. \]Practice 7: Independence
Suppose \(X\) is independent of event \(B\) and
\[ E[X]=12. \]Find \(E[X\mid B]\), assuming \(P(B)>0\).
Practice 8: Joint-Probability Formula
Suppose
\[ P(X=1\cap B)=0.10, \] \[ P(X=3\cap B)=0.20, \] \[ P(X=5\cap B)=0.10. \]Find \(E[X\mid B]\).
Practice 9: Conditional Profit
A project earns \(\$50{,}000\) under success and loses \(\$10{,}000\) under failure. Given information \(B\),
\[ P(\text{success}\mid B)=0.75. \]Find the conditional expected profit.
Practice 10: Identify the Error
A student knows
\[ E[X\mid B_1]=4 \]and
\[ E[X\mid B_2]=10 \]and concludes
\[ E[X]=7. \]Explain why this conclusion is not justified without additional information.
Practice Answers
Practice 1
Given \(X>3\), the possible outcomes are
\[ 4,5,6. \]Therefore,
\[ E[X\mid X>3] = \frac{4+5+6}{3} = \boxed{5}. \]Practice 2
\[ E[X\mid B] = 1(0.20)+2(0.50)+5(0.30). \]Therefore,
\[ E[X\mid B] = 0.20+1.00+1.50 = \boxed{2.70}. \]Practice 3
The conditional probabilities are
\[ 0.20,\qquad0.30,\qquad0.50. \]Thus,
\[ E[X\mid P] = 0(0.20)+1(0.30)+2(0.50) = \boxed{1.30}. \]Practice 4
\[ E[X] = 0.40(8)+0.60(3). \]Therefore,
\[ E[X] = 3.2+1.8 = \boxed{5}. \]Practice 5
\[ E[X] = 0.50(2)+0.30(5)+0.20(10). \]Therefore,
\[ E[X] = 1+1.5+2 = \boxed{4.5}. \]Practice 6
For an indicator variable,
\[ E[I_A\mid B]=P(A\mid B). \]Therefore,
\[ \boxed{ E[I_A\mid B]=0.65 }. \]Practice 7
Independence leaves the distribution of \(X\) unchanged under \(B\), so
\[ \boxed{ E[X\mid B]=12 }. \]Practice 8
First,
\[ P(B)=0.10+0.20+0.10=0.40. \]Then
\[ E[X\mid B] = \frac{ 1(0.10)+3(0.20)+5(0.10) }{0.40}. \]The numerator is
\[ 0.10+0.60+0.50=1.20. \]Therefore,
\[ \boxed{ E[X\mid B]=3 }. \]Practice 9
The failure probability given \(B\) is
\[ 0.25. \]Therefore,
\[ E[X\mid B] = 50000(0.75)+(-10000)(0.25). \]Thus,
\[ E[X\mid B] = 37500-2500 = \boxed{\$35{,}000}. \]Practice 10
The two conditional expectations must be weighted by the probabilities of their groups.
If \(B_1\) and \(B_2\) form a partition, then
\[ E[X] = P(B_1)E[X\mid B_1] + P(B_2)E[X\mid B_2]. \]Without the group probabilities, the overall expectation cannot be determined.
What to Remember
For a discrete random variable,
\[ \boxed{ E[X\mid B] = \sum_i x_iP(X=x_i\mid B) }. \]If
\[ B_1,\ldots,B_n \]form a partition, then
\[ \boxed{ E[X] = \sum_iP(B_i)E[X\mid B_i] }. \]For an indicator variable,
\[ \boxed{ E[I_A\mid B]=P(A\mid B) }. \]- Conditional expectation is an average under conditioned information.
- The weights are conditional probabilities.
- Conditioning can raise, lower, or leave unchanged an expected value.
- The law of total expectation combines subgroup expectations into an overall expectation.
- Conditional probability is a special case of conditional expectation through indicator variables.
- Equal conditional means do not imply equal conditional distributions.
- A conditional expectation is not necessarily an attainable outcome.
- In more advanced settings, \(E[X\mid Y]\) is itself a random variable.
The central habit is: update the probability weights first, then compute the average under the information that remains.
Where We Go Next
Conditional expectation summarizes how numerical outcomes change under known information. The next step is to represent dependence relationships explicitly so that complicated conditional structures can be organized and interpreted.
The next lesson, Dependency Diagrams, develops graphical representations of probabilistic dependence and conditional structure before the course moves into introductory Bayesian networks.