Question
Download Solution PDFIf x is a continuous random variable with p.d.f.
f(x) = 1/√2πe-x2/2dx , \( - ∞<x<∞\)
and y is defined as y = x + 1, then E(y) equals:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven
f(x) = 1/√2πe-x2/2dx
Formula
\(E\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty yf\left( x \right)dx\)
Calculation
\(E\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty yf\left( x \right)dx = \;\mathop \smallint \limits_{ - \infty }^\infty \left( {x + 1} \right)1/√ {2π } \)(e-x2/2dx
⇒ (1/√2π)\(\mathop \smallint \nolimits_{ - \infty }^\infty \left( {x + 1} \right)\)e-x2/2dx
⇒ E(y) = (1/√2π)\(\mathop \smallint \limits_{ - \infty }^\infty \)x × e-x2/2dx + (1/√2π)(\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx
⇒ Since, x × e-x2/2dx is an odd function of k
∴ (1/2√2π)(\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx = 0
⇒ E(y) = (1/2√2π)(\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx
⇒ (1/√2π) × √2π = 1
Using the standard integrals we have
\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx = √2π
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