If x is a continuous random variable with p.d.f.
f(x) = 1/√2πe-x2/2dx , $- \infty < x < \infty$

and y is defined as y = x + 1, then E(y) equals:

Question

Question

This question was previously asked in

SSC CGL Tier-II ( JSO ) 2016 Official Paper ( Held On : 2 Dec 2016 )

Answer 1

Given

f(x) = 1/√2πe^-x2^/2dx

Formula

$E (y) = \int_{- \infty}^{\infty} y f (x) d x$

Calculation

$E (y) = \int_{- \infty}^{\infty} y f (x) d x = \int_{- \infty}^{\infty} (x + 1) 1 / \sqrt 2 π$ (e^-x2/2dx

⇒ (1/√2π) $\int_{- \infty}^{\infty} (x + 1)$ e^-x2/2dx

⇒ E(y) = (1/√2π) $\int_{- \infty}^{\infty}$ x × e^-x2/2dx + (1/√2π)( $\int_{- \infty}^{\infty}$ e^-x2/2dx

⇒ Since, x × e^-x2/2^dx is an odd function of k

∴ (1/2√2π)( $\int_{- \infty}^{\infty}$ e^-x2/2dx = 0

⇒ E(y) = (1/2√2π)( $\int_{- \infty}^{\infty}$ e^-x2/2dx

⇒ (1/√2π) × √2π = 1

Using the standard integrals we have

$\int_{- \infty}^{\infty}$ e^-x2/2dx = √2π

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