if the probability density of x is given by
f(x)=2x^-3 for x>1, 0 elsewhere
check whether is mean and its variance exist
please steps by steps !
E(x)=[1,∞]∫xf(x)dx=[1,∞]∫x*2x^-3*dx=[1,∞]∫2x^-2*dx=[1,∞]{-2x^-1}=0-(-2)=2E(x^2)=[1,∞]∫x^ 2f (x)dx=[1,∞]∫x^2*2x^-3*dx=[1,∞]∫2x^-1dx=[1,∞]{2lnx}=∞-0=∞V(x)= E(x^2)- E(x)* E(x)= ∞Therefore, its mean exists but its variance does not exist.
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E(x)=[1,∞]∫xf(x)dx=[1,∞]∫x*2x^-3*dx=[1,∞]∫2x^-2*dx=[1,∞]{-2x^-1}=0-(-2)=2E(x^2)=[1,∞]∫x^ 2f (x)dx=[1,∞]∫x^2*2x^-3*dx=[1,∞]∫2x^-1dx=[1,∞]{2lnx}=∞-0=∞V(x)= E(x^2)- E(x)* E(x)= ∞Therefore, its mean exists but its variance does not exist.