G ghafsj New member Joined Dec 15, 2020 Messages 7 Dec 15, 2020 #1 . Let continuous random variable X have p.d.f. f(x) = λe−λx, x > 0, λ > 0 Find the p.d.f. of Y = 1/ X
. Let continuous random variable X have p.d.f. f(x) = λe−λx, x > 0, λ > 0 Find the p.d.f. of Y = 1/ X
R Romsek Senior Member Joined Nov 16, 2013 Messages 1,361 Dec 15, 2020 #2 I suppose you mean [MATH]f(x) = \lambda e^{-\lambda x}[/MATH], yes?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,603 Dec 15, 2020 #3 ghafsj said: . Let continuous random variable X have p.d.f. f(x) = λe−λx, x > 0, λ > 0 Find the p.d.f. of Y = 1/ X Click to expand... Please read the posting guidelines and follow them.
ghafsj said: . Let continuous random variable X have p.d.f. f(x) = λe−λx, x > 0, λ > 0 Find the p.d.f. of Y = 1/ X Click to expand... Please read the posting guidelines and follow them.
G ghafsj New member Joined Dec 15, 2020 Messages 7 Dec 15, 2020 #4 Romsek said: I suppose you mean [MATH]f(x) = \lambda e^{-\lambda x}[/MATH], yes? Click to expand... Yes
Romsek said: I suppose you mean [MATH]f(x) = \lambda e^{-\lambda x}[/MATH], yes? Click to expand... Yes
R Romsek Senior Member Joined Nov 16, 2013 Messages 1,361 Dec 15, 2020 #5 The procedure for finding the distribution of a function of a random variable [MATH]Y = f(X)[/MATH] is pretty standard. a) find [MATH]F_Y(y) = P[Y < y] = P[f(X) < y][/MATH] b) the PDF [MATH]f_Y(y) = \dfrac{d}{dy} F_Y(y)[/MATH]
The procedure for finding the distribution of a function of a random variable [MATH]Y = f(X)[/MATH] is pretty standard. a) find [MATH]F_Y(y) = P[Y < y] = P[f(X) < y][/MATH] b) the PDF [MATH]f_Y(y) = \dfrac{d}{dy} F_Y(y)[/MATH]
