We conclude then that var k 2 11 d chebyshevs

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We conclude then that Var( K )=2 1=1 . (d) Chebyshev’s inequality states that P ( | K E [ K ] | ≥ a ) Var(K) a 2 . Note that P ( K m )= P ( K 1 m 1) P ( | K 1 | ≥ m 1) . The last inequality comes from the fact that the event { K 1 m 1 } is included in the event {| K 1 | ≥ m 1 } . Applying Chebyshev’s inequality with a = m 1 we obtain P ( K m ) 1 ( m 1) 2 . 4. (a) We want P ( g ( X ) = k ) = P ( X = k ) for all integers k . Since g ( X ) = X , this is equivalent to p X ( k ) = P ( X = k )= P ( X = k )= p X ( k ) for all integers k . 5
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(b) If E [ X ] > 0 , then E [ g ( X )] = E [ X ] < 0 , consequently the random variables X and g ( X ) do not have the same mean and a fortiori they have different distributions. (c) Note that Var( aX ) = a 2 Var( X ) , hence Var( X ) = Var( X ) . As an example we can choose X such that P ( X = 1)= P ( X =1)= 1 2 , then X and X have the same distribution and so Var( X )=Var( X ) . (d) We want P (1 X = k )= P ( X = k ) for all integers k . This is equivalent to p X (1 k )= P ( X =1 k )= P ( X = k )= p X ( k ) for all integers k . (e) Select X to be a Bernoulli ( 1 2 ) random variable, then E [ X ]= 1 2 > 0 . The random variable 1 X is also going to be a Bernoulli ( 1 2 ) random variable. (f) Note that the relation Var(1 X )=Var( X ) is always true. Choosing X to be a Bernoulli ( 1 2 ) random variable works here as well as Var(1 X )=Var( X )= 1 4 > 0 and 1 X is also Bernoulli ( 1 2 ) . (g) We want P ( X 2 = k )= P ( X = k ) for all integers k , since X 2 0 we need to have p X ( k )= P ( X = k )=0 for all k < 0 . Now for k 0 , we must have P ( X = k )= P ( X = k ) or equivalently p X ( k )= p X ( k ) for all k 0 . (h) Select X to be Bernoulli ( p ) (with p> 0 ), then X 2 = X and we have E [ X 2 ]= E [ X ]= p> 0 . (i) Select X to be Bernoulli ( p ) (with 0 <p< 1 ), then X 2 = X and we have Var( X 2 )=Var( X )= p (1 p ) > 0 . 5. (a) The side of the square is 2 a . This means that its area is 2 a 2 and the joint PDF of X and Y is given by f XY ( x,y )= braceleftbigg 1 2 a 2 if ( x,y ) square 0 otherwise The PDF of X can be found by marginalizing the joint PDF of X and Y as follows f X ( x )= integraldisplay −∞ f XY ( x,y ) dy. (2) For a given x [0 ,a ] , we have f XY ( x,y ) is non zero only when y [ a + x,a x ] , which means that f X ( x )= integraldisplay a x a + x 1 2 a 2 dy = a x a 2 , x [0 ,a ] . (3) By symmetry we deduce f X ( x )= a + x a 2 , x [ a, 0] . (4) In conclusion we have f X ( x )= braceleftbigg a −| x | a 2 if a x a 0 otherwise (b) We start by computing the PDF of Y . We can directly use the result from part (a) and invoke symmetry to find f Y ( y )= braceleftbigg a −| y | a 2 if a y a 0 otherwise 6
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By definition, the conditional PDF of X given Y is given by f X | Y ( x | y )= braceleftBigg f XY ( x,y ) f Y ( y ) if a<y <a 0 otherwise For a given (fixed) a<y <a , this gives the following conditional PDF f X | Y ( x | y )= braceleftbigg 1 2( a −| y | ) if a + | y | ≤ x a − | y | 0 otherwise We can see then that given { Y = y } the conditional PDF of X is uniform on the interval [ a + | y | ,a − | y | ] .
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