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Question
Verify which of the following is p.d.f. of r.v. X:
f(x) = x, for 0 ≤ x ≤ 1 and 2 - x for 1 < x < 2
Solution
f (x) ≥ 0
`int_0^2 f(x) dx = int_0^1 f(x) dx + int_1^2 f(x) dx`
= `int_0^1 x dx + int_1^2 (2 - x) dx`
= `[x^2/2]_0^1 + [2x - x^2/2]_1^2 `
= `1^2/2 - 0^2/2 + (2 xx2 - 2^2/2) - (2 (1) - 1^2/2)`
= `1/2 - 0 + (4 - 2) - (2 - 1/2) `
= `1/2 + 2 - 3/2`
= `-2/2 + 2`
= - 1+ 2
= 1
∴ f (x) is p.d.f
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