HW04_2011 solution

# HW04_2011 solution - f ( x ) for &amp;amp;amp; &amp;amp;lt;...

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Problem Set 4 Solutions (1) A pmf must satisfy the following equation: P x 2 f X ( x ) = 1 P i c i = 1 : Since 1 P i 1 i = 1 , there is no possible value of c which makes P x 2 f X ( x ) equal to 1. (2) f ( x ) = 0 : 4 if t=1 = 0 : 2 if t=3 = 0 : 2 if t=5 = 0 : 2 if t=7 = 0 otherwise. (3) F X ( x ) = 0 if x < 0 = 0 : 41 if 0 x < 1 = 0 : 78 if 1 x < 2 = 0 : 94 if 2 x < 3 = 0 : 99 if 3 x < 4 = 1 if x > 4 (a) R 2 0 sin xdx = 1 : Thus, c = 1 : (b) R 1 e x j dx = R 0 e x dx + R 1 0 e x dx = 1 + 1 = 2 : Thus, c = 0 : 5 : (4) Since ± 1 < 1 + 2 sinx < 3 where ± & < x < &; there is no possible value of c ensuring that the pdf
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Unformatted text preview: f ( x ) for &amp; &lt; x &lt; &amp;: (5) F X ( x ) = 8 &gt; &gt; &gt; &gt; &gt; &lt; &gt; &gt; &gt; &gt; &gt; : ; x &lt; 1 R x &amp; 1 2 9 ( x + 1) dx = ( x +1) 2 9 ; 1 &amp; x &amp; 2 1 ;x &gt; 2 (6) P ( Y = y ) = P ( X X +1 = y ) = P ( X = y 1 &amp; y ) = 1 3 ( 2 3 ) y= (1 &amp; y ) ; where y = ; 1 2 ; 2 3 ; 3 4 ; ; x x +1 ; 1...
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