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प्रश्न
If X is the random variable with probability density function f(x) given by,
`f(x) = {{:(x + 1",", -1 ≤ x < 0),(-x +1",", 0 ≤ x < 1),(0, "otherwise"):}`
then find the distribution function F(x)
उत्तर
Distribution function F(x) = `int_-oo^x f(u) "d"u`
Case 1:
x < – 1
F(x)= `int_-oo^x f(u) "d"u` = 0
Case 2:
– 1 ≤ x <0
F(x) = `int_oo^x f(u) "d"u`
= `int_-oo^-1 f(u) "d"u + int_-1^x f(u) "d"u`
= `0 + int_-1^x (u + 1) "d"u`
= `["u"^2/2 + u]_-1^2`
= `x^2/2 + x - 1/2 + 1`
= `x^2/2 + x + 1/2`
Case 3:
0 ≤ x < 1
F(x) = `int_-oo^x f(u) "d"u`
= `int_oo^-1 f(u) "d"u + int_-1^0 f(u) "d"u + int_0^x f(u) "d"u`
= `0 + int_-1^0 (u + 1) "d"u + int_0^x (- u + 1) "d"u`
= `[u^2/2 + u]_-1^0 + [- u^2/2 + u]_0^x`
= `0 - (1/ - 1) + (- x^2/2 + x - 0)`
= `1/2 - x^2 + x`
Case 4:
x ≥ 1
F(x) = `int_oo^x f(u) "d"u`
= `int_-oo^-1 f(u) "d"u + int_-1^0 f(u) "d"u + int_0^1 f(u) "d"u + int_1^x f(u) "d"u`
= `0 + int_-1^0 (u + 1) du + int_0^1 (- u + 1) "d"u + 0`
= `[u^2/2 + u]_-1^0 + [- u^2/2 + u]_0^1`
= `0 -(1/2 - 1) + (- 1/2 + 1) + 1 - 0`
= `- 1/2 + 1 - 1/2 + 1`
F(x) = `{{:(0",", x ≤ - 1),(x^2/2 + x + 1/2",", - 1 ≤ x < 0),(1/2 - x^2/2 + x",", 0 ≤ x < 1),(1",", x ≥ 1):}`
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