Homogeneous distribution variance and correlation
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The random variable $X$ has a homogeneous distribution in the interval $(0,1)$ and $Y = X ^ k, k> 0.$ Calculate the variance and the correlation coefficient between the variables $X$ and $Y$.
probability
edited Dec 19 '18 at 22:39
dantopa
6,58942244
asked Dec 19 '18 at 18:57
AbakusAbakus
32
$endgroup$
add a comment |
$begingroup$
The random variable $X$ has a homogeneous distribution in the interval $(0,1)$ and $Y = X ^ k, k> 0.$ Calculate the variance and the correlation coefficient between the variables $X$ and $Y$.
probability
edited Dec 19 '18 at 22:39
dantopa
6,58942244
asked Dec 19 '18 at 18:57
AbakusAbakus
32
$endgroup$
add a comment |
$begingroup$
The random variable $X$ has a homogeneous distribution in the interval $(0,1)$ and $Y = X ^ k, k> 0.$ Calculate the variance and the correlation coefficient between the variables $X$ and $Y$.
probability
edited Dec 19 '18 at 22:39
dantopa
6,58942244
asked Dec 19 '18 at 18:57
AbakusAbakus
32
$endgroup$
The random variable $X$ has a homogeneous distribution in the interval $(0,1)$ and $Y = X ^ k, k> 0.$ Calculate the variance and the correlation coefficient between the variables $X$ and $Y$.
probability
probability
edited Dec 19 '18 at 22:39
dantopa
6,58942244
asked Dec 19 '18 at 18:57
AbakusAbakus
32
edited Dec 19 '18 at 22:39
dantopa
6,58942244
asked Dec 19 '18 at 18:57
AbakusAbakus
32
edited Dec 19 '18 at 22:39
dantopa
6,58942244
edited Dec 19 '18 at 22:39
dantopa
6,58942244
edited Dec 19 '18 at 22:39
dantopa
6,58942244
6,58942244
asked Dec 19 '18 at 18:57
AbakusAbakus
32
asked Dec 19 '18 at 18:57
AbakusAbakus
32
asked Dec 19 '18 at 18:57
AbakusAbakus
32
32
add a comment |
add a comment |
1 Answer
1
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$begingroup$
By some integration, we can find all these. Since this is a unit interval, the expected values are the "unweighted" integrals:
$$overline X=E[X]=int_0^1x,dx=frac12left[x^2right]_0^1=frac12$$
$$overline Y=E[Y]=int_0^1x^k,dx=frac1{k+1}left[x^{k+1}right]_0^1=frac1{k+1}$$
and the variances are
$$sigma_X^2=mathrm{Var}(X)=Eleft[(X-overline X)^2right]
=int_0^1left(x-tfrac12right)^2,dx=tfrac13left[left(x-tfrac12right)^3right]_0^1=tfrac1{12}$$
$$sigma_Y^2=mathrm{Var}(Y)=Eleft[(Y-overline Y)^2right]
=int_0^1left(x^k-tfrac1{k+1}right)^2,dx,$$
the latter of which can be found by
first multiplying out and then integrating,
or using $E[Y^2]-E[Y]^2$ then integrating,
and lastly simplifying to get
$$sigma_Y^2=frac{k^2}{(k+1)^2(2k+1)}.$$
Similarly, the covariance is
$$sigma_{XY}=mathrm{Cov}(X,Y)
=Eleft[(X-overline X)(Y-overline Y)right]
=E[XY]-E[X]E[Y]$$
$$=int_0^1x^{k+1},dx-tfrac1{2(k+1)}
=tfrac1{k+2}-tfrac1{2k+2}
=frac{k}{2(k+1)(k+2)}$$
so that the (Pearson) correlation coefficient is
$$
mathrm{Corr}(X,Y)
=frac{sigma_{XY}}{sigma_Xsigma_Y}
=frac{sqrt{3(2k+1)}}{k+2}.
$$
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By some integration, we can find all these. Since this is a unit interval, the expected values are the "unweighted" integrals:
$$overline X=E[X]=int_0^1x,dx=frac12left[x^2right]_0^1=frac12$$
$$overline Y=E[Y]=int_0^1x^k,dx=frac1{k+1}left[x^{k+1}right]_0^1=frac1{k+1}$$
and the variances are
$$sigma_X^2=mathrm{Var}(X)=Eleft[(X-overline X)^2right]
=int_0^1left(x-tfrac12right)^2,dx=tfrac13left[left(x-tfrac12right)^3right]_0^1=tfrac1{12}$$
$$sigma_Y^2=mathrm{Var}(Y)=Eleft[(Y-overline Y)^2right]
=int_0^1left(x^k-tfrac1{k+1}right)^2,dx,$$
the latter of which can be found by
first multiplying out and then integrating,
or using $E[Y^2]-E[Y]^2$ then integrating,
and lastly simplifying to get
$$sigma_Y^2=frac{k^2}{(k+1)^2(2k+1)}.$$
Similarly, the covariance is
$$sigma_{XY}=mathrm{Cov}(X,Y)
=Eleft[(X-overline X)(Y-overline Y)right]
=E[XY]-E[X]E[Y]$$
$$=int_0^1x^{k+1},dx-tfrac1{2(k+1)}
=tfrac1{k+2}-tfrac1{2k+2}
=frac{k}{2(k+1)(k+2)}$$
so that the (Pearson) correlation coefficient is
$$
mathrm{Corr}(X,Y)
=frac{sigma_{XY}}{sigma_Xsigma_Y}
=frac{sqrt{3(2k+1)}}{k+2}.
$$
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
$endgroup$
add a comment |
$begingroup$
By some integration, we can find all these. Since this is a unit interval, the expected values are the "unweighted" integrals:
$$overline X=E[X]=int_0^1x,dx=frac12left[x^2right]_0^1=frac12$$
$$overline Y=E[Y]=int_0^1x^k,dx=frac1{k+1}left[x^{k+1}right]_0^1=frac1{k+1}$$
and the variances are
$$sigma_X^2=mathrm{Var}(X)=Eleft[(X-overline X)^2right]
=int_0^1left(x-tfrac12right)^2,dx=tfrac13left[left(x-tfrac12right)^3right]_0^1=tfrac1{12}$$
$$sigma_Y^2=mathrm{Var}(Y)=Eleft[(Y-overline Y)^2right]
=int_0^1left(x^k-tfrac1{k+1}right)^2,dx,$$
the latter of which can be found by
first multiplying out and then integrating,
or using $E[Y^2]-E[Y]^2$ then integrating,
and lastly simplifying to get
$$sigma_Y^2=frac{k^2}{(k+1)^2(2k+1)}.$$
Similarly, the covariance is
$$sigma_{XY}=mathrm{Cov}(X,Y)
=Eleft[(X-overline X)(Y-overline Y)right]
=E[XY]-E[X]E[Y]$$
$$=int_0^1x^{k+1},dx-tfrac1{2(k+1)}
=tfrac1{k+2}-tfrac1{2k+2}
=frac{k}{2(k+1)(k+2)}$$
so that the (Pearson) correlation coefficient is
$$
mathrm{Corr}(X,Y)
=frac{sigma_{XY}}{sigma_Xsigma_Y}
=frac{sqrt{3(2k+1)}}{k+2}.
$$
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
$endgroup$
add a comment |
$begingroup$
By some integration, we can find all these. Since this is a unit interval, the expected values are the "unweighted" integrals:
$$overline X=E[X]=int_0^1x,dx=frac12left[x^2right]_0^1=frac12$$
$$overline Y=E[Y]=int_0^1x^k,dx=frac1{k+1}left[x^{k+1}right]_0^1=frac1{k+1}$$
and the variances are
$$sigma_X^2=mathrm{Var}(X)=Eleft[(X-overline X)^2right]
=int_0^1left(x-tfrac12right)^2,dx=tfrac13left[left(x-tfrac12right)^3right]_0^1=tfrac1{12}$$
$$sigma_Y^2=mathrm{Var}(Y)=Eleft[(Y-overline Y)^2right]
=int_0^1left(x^k-tfrac1{k+1}right)^2,dx,$$
the latter of which can be found by
first multiplying out and then integrating,
or using $E[Y^2]-E[Y]^2$ then integrating,
and lastly simplifying to get
$$sigma_Y^2=frac{k^2}{(k+1)^2(2k+1)}.$$
Similarly, the covariance is
$$sigma_{XY}=mathrm{Cov}(X,Y)
=Eleft[(X-overline X)(Y-overline Y)right]
=E[XY]-E[X]E[Y]$$
$$=int_0^1x^{k+1},dx-tfrac1{2(k+1)}
=tfrac1{k+2}-tfrac1{2k+2}
=frac{k}{2(k+1)(k+2)}$$
so that the (Pearson) correlation coefficient is
$$
mathrm{Corr}(X,Y)
=frac{sigma_{XY}}{sigma_Xsigma_Y}
=frac{sqrt{3(2k+1)}}{k+2}.
$$
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
$endgroup$
By some integration, we can find all these. Since this is a unit interval, the expected values are the "unweighted" integrals:
$$overline X=E[X]=int_0^1x,dx=frac12left[x^2right]_0^1=frac12$$
$$overline Y=E[Y]=int_0^1x^k,dx=frac1{k+1}left[x^{k+1}right]_0^1=frac1{k+1}$$
and the variances are
$$sigma_X^2=mathrm{Var}(X)=Eleft[(X-overline X)^2right]
=int_0^1left(x-tfrac12right)^2,dx=tfrac13left[left(x-tfrac12right)^3right]_0^1=tfrac1{12}$$
$$sigma_Y^2=mathrm{Var}(Y)=Eleft[(Y-overline Y)^2right]
=int_0^1left(x^k-tfrac1{k+1}right)^2,dx,$$
the latter of which can be found by
first multiplying out and then integrating,
or using $E[Y^2]-E[Y]^2$ then integrating,
and lastly simplifying to get
$$sigma_Y^2=frac{k^2}{(k+1)^2(2k+1)}.$$
Similarly, the covariance is
$$sigma_{XY}=mathrm{Cov}(X,Y)
=Eleft[(X-overline X)(Y-overline Y)right]
=E[XY]-E[X]E[Y]$$
$$=int_0^1x^{k+1},dx-tfrac1{2(k+1)}
=tfrac1{k+2}-tfrac1{2k+2}
=frac{k}{2(k+1)(k+2)}$$
so that the (Pearson) correlation coefficient is
$$
mathrm{Corr}(X,Y)
=frac{sigma_{XY}}{sigma_Xsigma_Y}
=frac{sqrt{3(2k+1)}}{k+2}.
$$
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
edited Dec 19 '18 at 23:47
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
answered Dec 19 '18 at 23:28
bginsbgins
7,5101322
7,5101322
add a comment |
add a comment |
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