35Expected value of
has density on . Find .
The mean is ∫x·f(x)dx over the support. With f(x)=2x on [0,1], that's ∫2x²dx from 0 to 1, giving [2x³/3] = 2/3. So E[X] = 2/3; and f checks out as a valid density since ∫2x dx = 1.
has density on . Find .
The mean is ∫x·f(x)dx over the support. With f(x)=2x on [0,1], that's ∫2x²dx from 0 to 1, giving [2x³/3] = 2/3. So E[X] = 2/3; and f checks out as a valid density since ∫2x dx = 1.