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प्रश्न
For the random variable X with the given probability mass function as below, find the mean and variance.
`f(x) = {{:(1/2 "e"^(x/2), "for" x > 0),(0, "otherwise"):}`
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उत्तर
Mean: `mu = "E"("X")`
= `int_0^oo x f(x) "d"x`
= `int_0^oo x 1/2 "e"^((-x)/2) "d"x`
= `1/2 [- 2x"e"^((-x)/2) ]_0^oo + int_0^oo 2"e"^((-x)/2) "d"x`
= `1/2[- 2x"e"^((-x)/2) + (2"e"^((-x)/2))/(- 1/2)]_0^oo`
= `1/2 [- 2x"e"^((-x)/2) - 4"e"^((-x)/2)]_0^oo`
= `1/2 [0 - 0 - (0 - 4)]`
= 2
Variance: `"E"("X"^2)`
= `int_0^oo x^2 f(x) "d"x`
`int u "dv" = "uv" - int "v" "du"`
u = `x int "dv" = int "e"^((-x)/2) "d"x`
du =dx, v = `"e"^((-x)/2)/(- 1/2)`
v = `- 2"e"^((-x)/2)`
Bernoulli's formula
`int"u" "dv" = "uv" - "u'v"_1 + "u''v"_2 - ......`
u = x2, `int"dv" = - int"e"^((-x)/2) "d"x`
u' = 2x, v = `- 2"e"^((-x)/2)`
u'' = 2, v1 = `- 4"e"^((-x)/2)`
u'' = 0, v2 = `- 8"e"^((-x)/2)`
`"E"("X"^2) = 1/2 int_0^oo x^2"e"^((-x)/2) "d"x`
= `1/2[-2x^2 "e"^((-x)/2) - 8x"e"^((-x)/2) - 16"e"^((-x)/2)]_0^oo`
= `1/2 [0 - (- 16"e"^0)]`
= `1/2 xx 16`
= 8
Var(X) = E(X2) – [E(X)]2
= 8 – 4
= 4
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