Question
Download Solution PDFयदि x प्रायिकता घनत्व फलन के साथ एक सतत यादृच्छिक चर है
f(x) = 1/√2πe-x2/2dx , \( - ∞<x<∞\)
और y को y = x + 1 के रूप में परिभाषित किया गया है, तो E(y) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है
f(x) = 1/√2πe-x2/2dx
सूत्र
\(E\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty yf\left( x \right)dx\)
गणना
\(E\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty yf\left( x \right)dx = \;\mathop \smallint \limits_{ - \infty }^\infty \left( {x + 1} \right)1/√ {2π } \)(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
⇒ चूंकि, x × e-x2/2dx k का विषम फलन है
∴ (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
मानक समाकलन का प्रयोग करने पर हमारे पास है
\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx = √2π
Last updated on Jul 9, 2025
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