given the following function: f(x)={( a e^(-a(x-b) ) a>0,x≥b 0 otherwise)┤ Prove that the function f(x) is a probability density function.
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total answers (1)
total answers (1)
Mathematics
It’s clear that: f(x)≥0 ∀x
need an explanation for this answer? contact us directly to get an explanation for this answer∫_(-∞)^∞〖f(x)dx=∫_(-∞)^∞〖a e^(-a(x-b) ) dx 〗〗
∫_(-∞)^a〖f(x)dx+∫_a^b〖f(x)dx 〗〗+∫_b^∞f(x)dx
=0+0+∫_b^∞〖a e^(-a(x-b) ) dx 〗
=a[e^(-a(x-b) )/(-a)]_b^∞=1
Thus the given function is probability density function.