Find the expectation of a random variable X if f(x) = ce⁻ ˣ for x>0 and 0 otherwise.
Solution:
∫0∞f(x)dx=1⇒∫0∞(ce−x)dx=1⇒−c[e−x]0∞=1⇒−c[0−1]=1⇒c=1\int_0^{\infty} f(x)dx=1 \\\Rightarrow\int_0^{\infty} (ce^{-x})dx=1 \\\Rightarrow -c[e^{-x}]_0^{\infty}=1 \\\Rightarrow -c[0-1]=1 \\\Rightarrow c=1∫0∞f(x)dx=1⇒∫0∞(ce−x)dx=1⇒−c[e−x]0∞=1⇒−c[0−1]=1⇒c=1
Then,
E[X]=∫0∞xf(x)dx=∫0∞x(ce−x)dx=∫0∞xe−xdx=[−e−xx−e−x]0∞=[0−0]−[0−1]=1E[X]=\int_0^{\infty}x f(x)dx \\=\int_0^{\infty}x (ce^{-x})dx \\=\int_0^{\infty}x e^{-x}dx \\=[-e^{-x}x-e^{-x}]_0^{\infty} \\=[0-0]-[0-1] \\=1E[X]=∫0∞xf(x)dx=∫0∞x(ce−x)dx=∫0∞xe−xdx=[−e−xx−e−x]0∞=[0−0]−[0−1]=1
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