Question

Solve the following problem :

Determine k if the p.d.f. of the r.v. is

f(x) = `{("ke"^(-thetax),  "for"  0 ≤ x < oo),(0, "otherwise".):}`
Find `"P"("X" > 1/theta)` and determine M is P(0 < X < M) = `(1)/(2)`

Answer 1

Since f(x) is the p.d.f. of X.

∴ `int_0^oo f(x)*dx` = 1

∴ `int_0^oo ("ke"^(-thetax))*dx` = 1

∴ `"k"*[("e"^(-thetax))/(-theta)]_0^oo` = 1

∴ `-"k"/theta ["e"^(-thetax))]_0^oo` = 1

∴ `- "K"/theta [1/"e"^(thetax)]_0^oo` = 1

∴ `-"k"/theta(1/oo - 1/1)` = 1

∴ `-"k"/theta(0 - 1)` = 1

∴ k = `theta`

F(x) = `"k" int_0^x "e"^(-thetax)*dx`

= `theta int_0^x e^(-thetax)*dx`        ...[∵ k = θ]

= `theta[("e"^(-thetax))/-theta]_0^x`

= `-["e"^(-thetax)]_0^x`

= `- ("e"^(-thetax) - 1)`

= `1 - (1)/"e"^(thetax)`                ...(i)

∴ `"P"("X" > 1/theta) = 1 - "P"("X" ≤ 1/theta)`

= `1 - "F"(1/theta)`

= `1 - [1 - 1/"e"^(theta(1/theta))]`      ...[From (i)]

= `(1)/"e"`
Given that, P(0 < X < M) = `(1)/(2)`

∴ F(M) – F(0) = `(1)/(2)`

∴ `1 - 1/"e"^(theta"M") - 0 = (1)/(2)`       ...[From (i)]

∴ `(1)/(2) = (1)/"e"^(theta"M")`

∴ e^θM = 2

∴ θM = log 2

∴ M = `(1)/theta log 2`.