Graduate · Conditional expectation by groups · 27 of 650
Conditional expectation by groups · First group baseline a=-0.5; Second group baseline b=-0.5
Use the completed group estimates at the animation endpoint. Find E[X|G] on both groups, E[X], and the mean squared prediction error. Givens: First group baseline a=-0.5; Second group baseline b=-0.5.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Four equally likely outcomes have X values (-0.5,1.5,-0.5,3.5). Information G reveals only whether the outcome is in the first pair or the second pair. Use the completed group estimates at the animation endpoint. Find E[X|G] on both groups, E[X], and the mean squared prediction error.
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A starting point
Average within each pair to get the two conditional values. Square all four residuals and average them with weight 1/4.
Work through the reasoning
Step 1
Identify the model and target
Four equally likely outcomes have X values (-0.5,1.5,-0.5,3.5). Information G reveals only whether the outcome is in the first pair or the second pair. Use the completed group estimates at the animation endpoint. The governing relation is X=(a,a+2,b,b+4); E[X|G]=(a+1,a+1,b+2,b+2). Average within each pair to get the two conditional values. Square all four residuals and average them with weight 1/4.
Step 2
Substitute and calculate
The exact group means are (-0.5+-0.5+2)/2=0.5 and (-0.5+-0.5+4)/2=1.5. At progress 1, interpolate each from overall mean 1, giving 0.5 and 1.5. Averaging all four squared residuals yields 2.5.
Step 3
Check the mathematical meaning
Residuals within the first pair are −1,+1 and within the second pair −2,+2. Thus MSE=(1+1+4+4)/4=2.5. Averaging the two group means gives E[X]=1. First group estimate: 0.5; Second group estimate: 1.5; Unconditional mean: 1; Mean squared error: 2.5. Decimal values are rounded, so use unrounded intermediate values.
The answer
The exact group means are (-0.5+-0.5+2)/2=0.5 and (-0.5+-0.5+4)/2=1.5. At progress 1, interpolate each from overall mean 1, giving 0.5 and 1.5. Averaging all four squared residuals yields 2.5. Residuals within the first pair are −1,+1 and within the second pair −2,+2. Thus MSE=(1+1+4+4)/4=2.5. Averaging the two group means gives E[X]=1. Animation check: First group estimate: 0.5; Second group estimate: 1.5; Unconditional mean: 1; Mean squared error: 2.5. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
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Watch the relationship
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The mathematical idea
Partial information may identify a group without revealing the exact outcome. Conditional expectation must be constant on each observable group and match that group's mean. Moving from the unconditional mean toward the two group means decreases squared error, illustrating the projection interpretation in a finite space. Four equally likely outcomes have X values (-0.5,1.5,-0.5,3.5). Information G reveals only whether the outcome is in the first pair or the second pair. Use the completed group estimates at the animation endpoint. The requested state occurs at 100% playback; use the exact target stated in the question for your calculation.
X=(a,a+2,b,b+4); E[X|G]=(a+1,a+1,b+2,b+2)
03 · Reflect and transfer
Explain what changes and why.
Why is E[X|G] usually a random quantity with two possible values, while E[X] is one scalar?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.