Let X be a continuous r.v has a p.f given by: f(x)={(ce^(-2x) , X≥0 0 otherwise)┤Find the value of c.
Since f(x) is a p.d.f, then ∫_(-∞)^∞〖f(x)〗 dx=1∴1=c∫_0^∞ e^(-2x) dx=c/2 [-e^(-2x) ]_0^∞=c/2c=2
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Since f(x) is a p.d.f, then ∫_(-∞)^∞〖f(x)〗 dx=1
need an explanation for this answer? contact us directly to get an explanation for this answer∴1=c∫_0^∞ e^(-2x) dx=c/2 [-e^(-2x) ]_0^∞=c/2
c=2