Undergraduate · Uniform convolution · 102 of 650
Uniform convolution · First uniform width a=1; Second uniform width b=2
Evaluate at sum value s=1.8, three fifths of the support length. Find the density f_S(s), cumulative probability P(S≤s), and expected sum. Givens: First uniform width a=1; Second uniform width b=2.
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
X and Y are independent continuous uniforms on [0,1] and [0,2], respectively. Let S=X+Y. Evaluate at sum value s=1.8, three fifths of the support length. Find the density f_S(s), cumulative probability P(S≤s), and expected sum.
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A starting point
Density equals the overlap length divided by ab. Integrate that piecewise-linear density to obtain a probability.
Work through the reasoning
Step 1
Identify the model and target
X and Y are independent continuous uniforms on [0,1] and [0,2], respectively. Let S=X+Y. Evaluate at sum value s=1.8, three fifths of the support length. The governing relation is f(x)=[x₊−(x−a)₊−(x−b)₊+(x−a−b)₊]/(ab). Density equals the overlap length divided by ab. Integrate that piecewise-linear density to obtain a probability.
Step 2
Substitute and calculate
At x=1.8, overlap length is max(0,min(x,a,b,a+b−x))=1. Divide by ab=(1)(2)=2 to get density 0.5. Integrating this piecewise-linear density through x gives cumulative probability 0.65.
Step 3
Check the mathematical meaning
The support is [0,3] and E[S]=1.5. The cumulative value 0.65 lies in [0,1]; density height 0.5 is not P(S=s), which is zero. Sum value: 1.8; Density at that value: 0.5; Cumulative probability: 0.65; Expected sum: 1.5. Decimal values are rounded, so use unrounded intermediate values.
The answer
At x=1.8, overlap length is max(0,min(x,a,b,a+b−x))=1. Divide by ab=(1)(2)=2 to get density 0.5. Integrating this piecewise-linear density through x gives cumulative probability 0.65. The support is [0,3] and E[S]=1.5. The cumulative value 0.65 lies in [0,1]; density height 0.5 is not P(S=s), which is zero. Animation check: Sum value: 1.8; Density at that value: 0.5; Cumulative probability: 0.65; Expected sum: 1.5. 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
The density of the sum of two independent uniforms is proportional to the overlap of two intervals. Equal widths give a triangle; unequal widths create a flat middle section. The density height is not a point probability, and integrating it gives the cumulative probability. X and Y are independent continuous uniforms on [0,1] and [0,2], respectively. Let S=X+Y. Evaluate at sum value s=1.8, three fifths of the support length. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
f(x)=[x₊−(x−a)₊−(x−b)₊+(x−a−b)₊]/(ab)
03 · Reflect and transfer
Explain what changes and why.
Why does swapping the two uniform widths leave the sum distribution unchanged?
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.