SOLUTION: A function f(x) has the following form: f(x)=kx^-(k+1) 1 < x < ∞ and zero otherwise. (a)For what values of k is f(x) a pdf? (b)Find the CDF based on (a). (c)For what val

Question 1203807: A function f(x) has the following form:
f(x)=kx^-(k+1) 1 < x < ∞
and zero otherwise.
(a)For what values of k is f(x) a pdf?
(b)Find the CDF based on (a).
(c)For what values of k does E(X) exist?
Answer by amarjeeth123(569) (Show Source): You can put this solution on YOUR website!
f(x)={kx−(k+1)01 a)
∫−∞∞f(x)dx=1∫−∞∞f(x)dx=1

Then
k≠0k=0
∫1∞kx−(k+1)dx=lim⁡A→∞[k−(k+1)+1x−(k+1)+1]A1∫1∞kx−(k+1)dx=A→∞lim[−(k+1)+1kx−(k+1)+1]A1
=−lim⁡A→∞A−k+1=1=−A→∞limA−k+1=1
−lim⁡A→∞A−k=0=>k>0−A→∞limA−k=0=>k>0
Answer: k>0k>0
b)

F(x)=∫−∞xf(y)dyF(x)=∫−∞xf(y)dy
F(x)=0,x<1F(x)=0,x<1
F(x)=∫1xky−(k+1)dyF(x)=∫1xky−(k+1)dy
=[k−(k+1)+1y−(k+1)+1]x1=−x−k+1=[−(k+1)+1ky−(k+1)+1]x1=−x−k+1

Answer:
F(x)={0x<11−x−k1≤x<∞,k>0F(x)={01−x−kx<11≤x<∞,k>0
c)

E(X)=∫−∞∞xf(x)dx=∫1∞kx−(k+1)+1dxE(X)=∫−∞∞xf(x)dx=∫1∞kx−(k+1)+1dx
=lim⁡A→∞[k−(k+1)+2x−(k+1)+2]A1=A→∞lim[−(k+1)+2kx−(k+1)+2]A1
=−kk−1lim⁡A→∞A−k+1+kk−1,k>1=−k−1kA→∞limA−k+1+k−1k,k>1
If k=1k=1

E(X)=∫−∞∞xf(x)dx=∫1∞kx−1dxE(X)=∫−∞∞xf(x)dx=∫1∞kx−1dx
=lim⁡A→∞[kln⁡(∣x∣)]A1=does not exist=A→∞lim[kln(∣x∣)]A1=does not exist

Answer:
E(X)E(X) exists for k>1.k>1.
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