Random Variables and Probability
For each of the conditions (a)-(d) below, give examples of random variables that satisfy the condition, and justify. If no such random variables exist, say so and provide justification.
a) Random variables X, Y such that Var (X + Y ) < Var X + Var Y
b) Random variables X, Y such that E(X|Y = y) > E(X) for all y
c) Independent random variables X, Y such that X+Y/Sqrt(2) has the same pdf
as X
d) Random variables Xi that are i.i.d, with mean and variance equal to 1,
such that P( Summation of Xi > 2n, for i=1 to n) tends to 0.5 as n tends
to become infinite
I have solved (a) and it would be true for negative covariance. How to solve the rest. Any help would be appreciated.
probability-distributions random-variables
asked Dec 1 at 11:50
Robo4.4
33
|
show 2 more comments
For each of the conditions (a)-(d) below, give examples of random variables that satisfy the condition, and justify. If no such random variables exist, say so and provide justification.
a) Random variables X, Y such that Var (X + Y ) < Var X + Var Y
b) Random variables X, Y such that E(X|Y = y) > E(X) for all y
c) Independent random variables X, Y such that X+Y/Sqrt(2) has the same pdf
as X
d) Random variables Xi that are i.i.d, with mean and variance equal to 1,
such that P( Summation of Xi > 2n, for i=1 to n) tends to 0.5 as n tends
to become infinite
I have solved (a) and it would be true for negative covariance. How to solve the rest. Any help would be appreciated.
probability-distributions random-variables
asked Dec 1 at 11:50
Robo4.4
33
Image not visible.
– Kavi Rama Murthy
Dec 1 at 11:59
b) is not possible since it implies $EX>EX$. For c) take i.i.d. standard normal.
– Kavi Rama Murthy
Dec 1 at 12:01
ok let me check
– Robo4.4
Dec 1 at 12:01
How for b) EX>EX and for c) what u mean by take iid standard normal... can you show the working?
– Robo4.4
Dec 1 at 12:02
Not allowed to embed image.
– Robo4.4
Dec 1 at 12:07
|
show 2 more comments
For each of the conditions (a)-(d) below, give examples of random variables that satisfy the condition, and justify. If no such random variables exist, say so and provide justification.
a) Random variables X, Y such that Var (X + Y ) < Var X + Var Y
b) Random variables X, Y such that E(X|Y = y) > E(X) for all y
c) Independent random variables X, Y such that X+Y/Sqrt(2) has the same pdf
as X
d) Random variables Xi that are i.i.d, with mean and variance equal to 1,
such that P( Summation of Xi > 2n, for i=1 to n) tends to 0.5 as n tends
to become infinite
I have solved (a) and it would be true for negative covariance. How to solve the rest. Any help would be appreciated.
probability-distributions random-variables
asked Dec 1 at 11:50
Robo4.4
33
For each of the conditions (a)-(d) below, give examples of random variables that satisfy the condition, and justify. If no such random variables exist, say so and provide justification.
a) Random variables X, Y such that Var (X + Y ) < Var X + Var Y
b) Random variables X, Y such that E(X|Y = y) > E(X) for all y
c) Independent random variables X, Y such that X+Y/Sqrt(2) has the same pdf
as X
d) Random variables Xi that are i.i.d, with mean and variance equal to 1,
such that P( Summation of Xi > 2n, for i=1 to n) tends to 0.5 as n tends
to become infinite
I have solved (a) and it would be true for negative covariance. How to solve the rest. Any help would be appreciated.
probability-distributions random-variables
probability-distributions random-variables
asked Dec 1 at 11:50
Robo4.4
33
asked Dec 1 at 11:50
Robo4.4
33
edited Dec 1 at 12:03
asked Dec 1 at 11:50
Robo4.4
33
asked Dec 1 at 11:50
Robo4.4
33
asked Dec 1 at 11:50
Robo4.4
33
33
Image not visible.
– Kavi Rama Murthy
Dec 1 at 11:59
b) is not possible since it implies $EX>EX$. For c) take i.i.d. standard normal.
– Kavi Rama Murthy
Dec 1 at 12:01
ok let me check
– Robo4.4
Dec 1 at 12:01
How for b) EX>EX and for c) what u mean by take iid standard normal... can you show the working?
– Robo4.4
Dec 1 at 12:02
Not allowed to embed image.
– Robo4.4
Dec 1 at 12:07
|
show 2 more comments
Image not visible.
– Kavi Rama Murthy
Dec 1 at 11:59
b) is not possible since it implies $EX>EX$. For c) take i.i.d. standard normal.
– Kavi Rama Murthy
Dec 1 at 12:01
ok let me check
– Robo4.4
Dec 1 at 12:01
How for b) EX>EX and for c) what u mean by take iid standard normal... can you show the working?
– Robo4.4
Dec 1 at 12:02
Not allowed to embed image.
– Robo4.4
Dec 1 at 12:07
Image not visible.
– Kavi Rama Murthy
Dec 1 at 11:59
Image not visible.
– Kavi Rama Murthy
Dec 1 at 11:59
b) is not possible since it implies $EX>EX$. For c) take i.i.d. standard normal.
– Kavi Rama Murthy
Dec 1 at 12:01
b) is not possible since it implies $EX>EX$. For c) take i.i.d. standard normal.
– Kavi Rama Murthy
Dec 1 at 12:01
ok let me check
– Robo4.4
Dec 1 at 12:01
ok let me check
– Robo4.4
Dec 1 at 12:01
How for b) EX>EX and for c) what u mean by take iid standard normal... can you show the working?
– Robo4.4
Dec 1 at 12:02
How for b) EX>EX and for c) what u mean by take iid standard normal... can you show the working?
– Robo4.4
Dec 1 at 12:02
Not allowed to embed image.
– Robo4.4
Dec 1 at 12:07
Not allowed to embed image.
– Robo4.4
Dec 1 at 12:07
|
show 2 more comments
1 Answer
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active
oldest
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b) says $EXI_{Y=y} >EX P{Y=y}$ in the disecrete case. Summing over $y$ we get $EX>EX$ In general case use the fact that $E(EX|Y))=EX$ to get the contradiction $EX >EX$. So b) is not possible. For c) take $X,Y$ i.i.d. standard normal and use the fact that characteristic function of standard normal distribution is $e^{-t^{2}/2}$. Using this you can easily check that $frac {X+Y} {sqrt 2}$ also has the same characteristic function. [$Ee^{itfrac {X+Y} {sqrt 2}}=Ee^{itfrac X {sqrt 2}}Ee^{itfrac Y {sqrt 2}}=e^{-t^{2}/4}e^{-t^{2}/4}=e^{-t^{2}/2}$].
d) is not possible by SLLN: since $frac 1 n sum_{i=1}^{n} X_i to 1$ almost surely the limit is necessarily $0$.
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
add a comment |
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b) says $EXI_{Y=y} >EX P{Y=y}$ in the disecrete case. Summing over $y$ we get $EX>EX$ In general case use the fact that $E(EX|Y))=EX$ to get the contradiction $EX >EX$. So b) is not possible. For c) take $X,Y$ i.i.d. standard normal and use the fact that characteristic function of standard normal distribution is $e^{-t^{2}/2}$. Using this you can easily check that $frac {X+Y} {sqrt 2}$ also has the same characteristic function. [$Ee^{itfrac {X+Y} {sqrt 2}}=Ee^{itfrac X {sqrt 2}}Ee^{itfrac Y {sqrt 2}}=e^{-t^{2}/4}e^{-t^{2}/4}=e^{-t^{2}/2}$].
d) is not possible by SLLN: since $frac 1 n sum_{i=1}^{n} X_i to 1$ almost surely the limit is necessarily $0$.
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
add a comment |
b) says $EXI_{Y=y} >EX P{Y=y}$ in the disecrete case. Summing over $y$ we get $EX>EX$ In general case use the fact that $E(EX|Y))=EX$ to get the contradiction $EX >EX$. So b) is not possible. For c) take $X,Y$ i.i.d. standard normal and use the fact that characteristic function of standard normal distribution is $e^{-t^{2}/2}$. Using this you can easily check that $frac {X+Y} {sqrt 2}$ also has the same characteristic function. [$Ee^{itfrac {X+Y} {sqrt 2}}=Ee^{itfrac X {sqrt 2}}Ee^{itfrac Y {sqrt 2}}=e^{-t^{2}/4}e^{-t^{2}/4}=e^{-t^{2}/2}$].
d) is not possible by SLLN: since $frac 1 n sum_{i=1}^{n} X_i to 1$ almost surely the limit is necessarily $0$.
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
add a comment |
b) says $EXI_{Y=y} >EX P{Y=y}$ in the disecrete case. Summing over $y$ we get $EX>EX$ In general case use the fact that $E(EX|Y))=EX$ to get the contradiction $EX >EX$. So b) is not possible. For c) take $X,Y$ i.i.d. standard normal and use the fact that characteristic function of standard normal distribution is $e^{-t^{2}/2}$. Using this you can easily check that $frac {X+Y} {sqrt 2}$ also has the same characteristic function. [$Ee^{itfrac {X+Y} {sqrt 2}}=Ee^{itfrac X {sqrt 2}}Ee^{itfrac Y {sqrt 2}}=e^{-t^{2}/4}e^{-t^{2}/4}=e^{-t^{2}/2}$].
d) is not possible by SLLN: since $frac 1 n sum_{i=1}^{n} X_i to 1$ almost surely the limit is necessarily $0$.
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
b) says $EXI_{Y=y} >EX P{Y=y}$ in the disecrete case. Summing over $y$ we get $EX>EX$ In general case use the fact that $E(EX|Y))=EX$ to get the contradiction $EX >EX$. So b) is not possible. For c) take $X,Y$ i.i.d. standard normal and use the fact that characteristic function of standard normal distribution is $e^{-t^{2}/2}$. Using this you can easily check that $frac {X+Y} {sqrt 2}$ also has the same characteristic function. [$Ee^{itfrac {X+Y} {sqrt 2}}=Ee^{itfrac X {sqrt 2}}Ee^{itfrac Y {sqrt 2}}=e^{-t^{2}/4}e^{-t^{2}/4}=e^{-t^{2}/2}$].
d) is not possible by SLLN: since $frac 1 n sum_{i=1}^{n} X_i to 1$ almost surely the limit is necessarily $0$.
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
edited Dec 1 at 23:17
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
answered Dec 1 at 12:19
Kavi Rama Murthy
49.2k31854
49.2k31854
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
add a comment |
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
Not sure of c) do we need to use e−t2/2 separately for X and Y and then add it and divide by root(2). How should I prove it.
– Robo4.4
Dec 1 at 13:24
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
@Robo4.4 I have added one more step to the answer. The first equality is by independence. In the second equality I have replaced $t$ by $t/sqrt 2$ in the gaussian characteristic function.
– Kavi Rama Murthy
Dec 1 at 23:18
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
In the second equality you mean to say you have replaced X by t/root(2)?
– Robo4.4
Dec 2 at 5:28
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
@Robo4.4 Write down the formula $Ee^{itx}=e^{-t^{2}/2}$ valid for all $t$ and replace $t$ by $t/sqrt 2$.
– Kavi Rama Murthy
Dec 2 at 11:30
add a comment |
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– Kavi Rama Murthy
Dec 1 at 11:59
b) is not possible since it implies $EX>EX$. For c) take i.i.d. standard normal.
– Kavi Rama Murthy
Dec 1 at 12:01
ok let me check
– Robo4.4
Dec 1 at 12:01
How for b) EX>EX and for c) what u mean by take iid standard normal... can you show the working?
– Robo4.4
Dec 1 at 12:02
Not allowed to embed image.
– Robo4.4
Dec 1 at 12:07