Probability — Question 5

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Question 5

Two densities on [0,1][0,1] are f(x)=2xf(x)=2x and g(x)=2(1−x)g(x)=2(1-x). Compare their means without integration by identifying a reflection.

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Original worksheet page 1: question and worked solution for 2-5-005
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Question 5 – Solution

See the diagram in the original worksheet below.

Step 1: Identify the reflection. f(1−x)=2(1−x)=g(x).f(1-x)=2(1-x)=g(x). Therefore, the graph of gg is the reflection of the graph of ff across x=1/2x=1/2.

Step 2: Find the mean of ff geometrically. The area under f(x)=2xf(x)=2x is a right triangle with vertices (0,0),(1,0),(1,2).(0,0),\qquad (1,0),\qquad (1,2). The xx-coordinate of a triangle’s centroid is the average of its vertex xx-coordinates: μf=0+1+13=23.\mu_f=\frac{0+1+1}{3}=\frac23.

Step 3: Reflect the mean. Reflection across x=1/2x=1/2 sends a location xx to 1−x1-x. Hence μg=1−μf=1−23=13.\mu_g=1-\mu_f=1-\frac23=\frac13. μf=23,μg=13\boxed{\mu_f=\frac23,\qquad \mu_g=\frac13} As a check, the two reflected means add to 11.

Original worksheet page 2: question and worked solution for 2-5-005

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