Mixing
If $X,Y sim chi ^2$ with $m$ and $n$ degrees of freedom, then $frac{X}{X+Y} = beta(m/2, n/2)$
I was reading https://en.wikipedia.org/wiki/Ratio_distribution#Other_ratio_distributions and I find the following problem:
Let $X$ and $Y$ be $chi ^2 $ with $m$ and $n$ degrees of freedom then $dfrac{X}{X+Y} = beta(m/2, n/2)$. Where $chi^2(k)$ has density function $$f_{chi^2(k)}(x)=begin{cases}displaystyle
frac{1}{2^{k/2}Gamma(k/2)},x^{(k/2) - 1} e^{-x/2}&text{if }x>0,\
0&text{if }xle0
end{cases} $$
I proved that $X+Y$ is $chi ^2$ with $n + m$ degrees of freedom using the moment-generating function. If $Z=X+Y$ and $U=dfrac{X}{X+Y}=dfrac{X}{Z}$ using the formula for the density of the ratio of two independent random variables:
$$f_U(u)= int_{-infty}^{infty}f_X(uv)f_Z(v)|v|dv $$
then
$$dfrac{u^{m/2-1}}{2^{(m+n/2)} Gamma(m/2) Gamma((m+n)/2)} int_0^{infty} v^{m+dfrac{n}{2}-1} e^{-(v/2)(u+1)}dv $$
and using some algebra:
$$f_U(u)=dfrac{u^{m/2-1} Gamma(m+n/2)}{Gamma(m+n/2) Gamma(m/2) (u+1)^{m+n/2}}$$
and this is not $beta(m/2, n/2)$, it looks more like $beta(m/2, m/2+n/2)$ but $(u+1)^{m+n/2}$ is causing troubles.
I need help if something is false or any hint.
probability probability-distributions random-variables density-function
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
asked May 3 '17 at 7:50
Mike A.
567414
add a comment |
I was reading https://en.wikipedia.org/wiki/Ratio_distribution#Other_ratio_distributions and I find the following problem:
Let $X$ and $Y$ be $chi ^2 $ with $m$ and $n$ degrees of freedom then $dfrac{X}{X+Y} = beta(m/2, n/2)$. Where $chi^2(k)$ has density function $$f_{chi^2(k)}(x)=begin{cases}displaystyle
frac{1}{2^{k/2}Gamma(k/2)},x^{(k/2) - 1} e^{-x/2}&text{if }x>0,\
0&text{if }xle0
end{cases} $$
I proved that $X+Y$ is $chi ^2$ with $n + m$ degrees of freedom using the moment-generating function. If $Z=X+Y$ and $U=dfrac{X}{X+Y}=dfrac{X}{Z}$ using the formula for the density of the ratio of two independent random variables:
$$f_U(u)= int_{-infty}^{infty}f_X(uv)f_Z(v)|v|dv $$
then
$$dfrac{u^{m/2-1}}{2^{(m+n/2)} Gamma(m/2) Gamma((m+n)/2)} int_0^{infty} v^{m+dfrac{n}{2}-1} e^{-(v/2)(u+1)}dv $$
and using some algebra:
$$f_U(u)=dfrac{u^{m/2-1} Gamma(m+n/2)}{Gamma(m+n/2) Gamma(m/2) (u+1)^{m+n/2}}$$
and this is not $beta(m/2, n/2)$, it looks more like $beta(m/2, m/2+n/2)$ but $(u+1)^{m+n/2}$ is causing troubles.
I need help if something is false or any hint.
probability probability-distributions random-variables density-function
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
asked May 3 '17 at 7:50
Mike A.
567414
It is better to give the expression of $chi ^2 $, might help for the reader.
– Frey
May 3 '17 at 7:56
4
I'm pretty sure $X$ and $Z$ are not independent.
– Em.
May 3 '17 at 7:58
Thank you, the mistake was to consider $X$ and $Z$ independent
– Mike A.
May 3 '17 at 9:36
add a comment |
I was reading https://en.wikipedia.org/wiki/Ratio_distribution#Other_ratio_distributions and I find the following problem:
Let $X$ and $Y$ be $chi ^2 $ with $m$ and $n$ degrees of freedom then $dfrac{X}{X+Y} = beta(m/2, n/2)$. Where $chi^2(k)$ has density function $$f_{chi^2(k)}(x)=begin{cases}displaystyle
frac{1}{2^{k/2}Gamma(k/2)},x^{(k/2) - 1} e^{-x/2}&text{if }x>0,\
0&text{if }xle0
end{cases} $$
I proved that $X+Y$ is $chi ^2$ with $n + m$ degrees of freedom using the moment-generating function. If $Z=X+Y$ and $U=dfrac{X}{X+Y}=dfrac{X}{Z}$ using the formula for the density of the ratio of two independent random variables:
$$f_U(u)= int_{-infty}^{infty}f_X(uv)f_Z(v)|v|dv $$
then
$$dfrac{u^{m/2-1}}{2^{(m+n/2)} Gamma(m/2) Gamma((m+n)/2)} int_0^{infty} v^{m+dfrac{n}{2}-1} e^{-(v/2)(u+1)}dv $$
and using some algebra:
$$f_U(u)=dfrac{u^{m/2-1} Gamma(m+n/2)}{Gamma(m+n/2) Gamma(m/2) (u+1)^{m+n/2}}$$
and this is not $beta(m/2, n/2)$, it looks more like $beta(m/2, m/2+n/2)$ but $(u+1)^{m+n/2}$ is causing troubles.
I need help if something is false or any hint.
probability probability-distributions random-variables density-function
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
asked May 3 '17 at 7:50
Mike A.
567414
I was reading https://en.wikipedia.org/wiki/Ratio_distribution#Other_ratio_distributions and I find the following problem:
Let $X$ and $Y$ be $chi ^2 $ with $m$ and $n$ degrees of freedom then $dfrac{X}{X+Y} = beta(m/2, n/2)$. Where $chi^2(k)$ has density function $$f_{chi^2(k)}(x)=begin{cases}displaystyle
frac{1}{2^{k/2}Gamma(k/2)},x^{(k/2) - 1} e^{-x/2}&text{if }x>0,\
0&text{if }xle0
end{cases} $$
I proved that $X+Y$ is $chi ^2$ with $n + m$ degrees of freedom using the moment-generating function. If $Z=X+Y$ and $U=dfrac{X}{X+Y}=dfrac{X}{Z}$ using the formula for the density of the ratio of two independent random variables:
$$f_U(u)= int_{-infty}^{infty}f_X(uv)f_Z(v)|v|dv $$
then
$$dfrac{u^{m/2-1}}{2^{(m+n/2)} Gamma(m/2) Gamma((m+n)/2)} int_0^{infty} v^{m+dfrac{n}{2}-1} e^{-(v/2)(u+1)}dv $$
and using some algebra:
$$f_U(u)=dfrac{u^{m/2-1} Gamma(m+n/2)}{Gamma(m+n/2) Gamma(m/2) (u+1)^{m+n/2}}$$
and this is not $beta(m/2, n/2)$, it looks more like $beta(m/2, m/2+n/2)$ but $(u+1)^{m+n/2}$ is causing troubles.
I need help if something is false or any hint.
probability probability-distributions random-variables density-function
probability probability-distributions random-variables density-function
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
asked May 3 '17 at 7:50
Mike A.
567414
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
asked May 3 '17 at 7:50
Mike A.
567414
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
edited Nov 20 '18 at 15:59
StubbornAtom
5,30711138
5,30711138
asked May 3 '17 at 7:50
Mike A.
567414
asked May 3 '17 at 7:50
Mike A.
567414
asked May 3 '17 at 7:50
Mike A.
567414
567414
It is better to give the expression of $chi ^2 $, might help for the reader.
– Frey
May 3 '17 at 7:56
4
I'm pretty sure $X$ and $Z$ are not independent.
– Em.
May 3 '17 at 7:58
Thank you, the mistake was to consider $X$ and $Z$ independent
– Mike A.
May 3 '17 at 9:36
add a comment |
It is better to give the expression of $chi ^2 $, might help for the reader.
– Frey
May 3 '17 at 7:56
4
I'm pretty sure $X$ and $Z$ are not independent.
– Em.
May 3 '17 at 7:58
Thank you, the mistake was to consider $X$ and $Z$ independent
– Mike A.
May 3 '17 at 9:36
It is better to give the expression of $chi ^2 $, might help for the reader.
– Frey
May 3 '17 at 7:56
It is better to give the expression of $chi ^2 $, might help for the reader.
– Frey
May 3 '17 at 7:56
4
4
I'm pretty sure $X$ and $Z$ are not independent.
– Em.
May 3 '17 at 7:58
I'm pretty sure $X$ and $Z$ are not independent.
– Em.
May 3 '17 at 7:58
Thank you, the mistake was to consider $X$ and $Z$ independent
– Mike A.
May 3 '17 at 9:36
Thank you, the mistake was to consider $X$ and $Z$ independent
– Mike A.
May 3 '17 at 9:36
add a comment |
2 Answers
2
active
oldest
votes
I may start as follows, then you do not need to worry about the dependency:
begin{equation}
label{eq_cd}
begin{split}
F_{U}(u)
=&{mathbb P}left[frac{X}{X+Y} leq uright] ={mathbb P}left[Ygeq frac{(1-u)X}{u}right] \
=& 1-int_{0}^{infty}F_Yleft(frac{(1-u)x}{u}right)f_{X}(x) dx \
end{split}
end{equation}
Since you know both CDF and PDF of $chi^2$, you can evaluate $F_{U}(u)$ which may be with $_2 F_1(.,.,.,.)$ funtion. However, you may write $_2 F_1(.,.,.,.)$ with $B(.,.)$ function, if you really need it.
Then, you can find $f_{U}(u)=frac{d }{du}F_{U}(u)$.
answered May 3 '17 at 8:24
Frey
648312
add a comment |
Let $X sim chi^2(m)$ and $Y sim chi^2(n)$, and let $T(X,Y)=(U,V)$ be a transformation such that $T_1(X,Y)=U=frac{X}{X+Y}$ and $T_2(X,Y)=V=X+Y$, then $T^{-1}_1(U,V)=X= UV$ and $T^{-1}_2(U,V)=Y=V-UV$. Then $ |J_{T^{-1}}|=|V| $. $X$ and $Y$ are independent then the joint density function is:
$$ f_{X,Y}(x,y)=frac{(1/2)^{m/2}}{Gamma(m/2)}x^{m/2 -1} e^{-x/2} frac{(1/2)^{n/2}}{Gamma(n/2)}y^{n/2 -1} e^{-y/2} $$
for $x, y geq 0$ and $0$ in other case.
Then, by the theorem of change of variable we have:
$$f_U(u)= int_{-infty}^{infty} frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} (uv)^{m/2-1}(v-uv)^{n/2-1}e^{-uv/2-(v-uv)/2} |v|dv $$
We have $v=x+y$ then $f_U$ is $0$ for $v < 0$ then we can consider the integral only in the interval $(0, infty)$ and $|v|=v$, also $u=frac{x}{x+y}$ then $0 leq u leq 1$. With some algebra:
$$f_U(u)= frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv $$
If $w=v/2$ and we make the change of variable in the integral we obtain:
$$int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv = 2^{m/2+n/2} Gamma(m/2+n/2)$$
Hence,
$$f_U(u)=frac{ Gamma(m/2+n/2) }{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} =dfrac{u^{m/2-1} (1-u)^{n/2-1}}{B(m/2,n/2)}$$
in the interval $[0,1]$, and this is the density function of the Beta distribution.
answered May 3 '17 at 10:26
Mike A.
567414
add a comment |
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2 Answers
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2 Answers
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I may start as follows, then you do not need to worry about the dependency:
begin{equation}
label{eq_cd}
begin{split}
F_{U}(u)
=&{mathbb P}left[frac{X}{X+Y} leq uright] ={mathbb P}left[Ygeq frac{(1-u)X}{u}right] \
=& 1-int_{0}^{infty}F_Yleft(frac{(1-u)x}{u}right)f_{X}(x) dx \
end{split}
end{equation}
Since you know both CDF and PDF of $chi^2$, you can evaluate $F_{U}(u)$ which may be with $_2 F_1(.,.,.,.)$ funtion. However, you may write $_2 F_1(.,.,.,.)$ with $B(.,.)$ function, if you really need it.
Then, you can find $f_{U}(u)=frac{d }{du}F_{U}(u)$.
answered May 3 '17 at 8:24
Frey
648312
add a comment |
I may start as follows, then you do not need to worry about the dependency:
begin{equation}
label{eq_cd}
begin{split}
F_{U}(u)
=&{mathbb P}left[frac{X}{X+Y} leq uright] ={mathbb P}left[Ygeq frac{(1-u)X}{u}right] \
=& 1-int_{0}^{infty}F_Yleft(frac{(1-u)x}{u}right)f_{X}(x) dx \
end{split}
end{equation}
Since you know both CDF and PDF of $chi^2$, you can evaluate $F_{U}(u)$ which may be with $_2 F_1(.,.,.,.)$ funtion. However, you may write $_2 F_1(.,.,.,.)$ with $B(.,.)$ function, if you really need it.
Then, you can find $f_{U}(u)=frac{d }{du}F_{U}(u)$.
answered May 3 '17 at 8:24
Frey
648312
add a comment |
I may start as follows, then you do not need to worry about the dependency:
begin{equation}
label{eq_cd}
begin{split}
F_{U}(u)
=&{mathbb P}left[frac{X}{X+Y} leq uright] ={mathbb P}left[Ygeq frac{(1-u)X}{u}right] \
=& 1-int_{0}^{infty}F_Yleft(frac{(1-u)x}{u}right)f_{X}(x) dx \
end{split}
end{equation}
Since you know both CDF and PDF of $chi^2$, you can evaluate $F_{U}(u)$ which may be with $_2 F_1(.,.,.,.)$ funtion. However, you may write $_2 F_1(.,.,.,.)$ with $B(.,.)$ function, if you really need it.
Then, you can find $f_{U}(u)=frac{d }{du}F_{U}(u)$.
answered May 3 '17 at 8:24
Frey
648312
I may start as follows, then you do not need to worry about the dependency:
begin{equation}
label{eq_cd}
begin{split}
F_{U}(u)
=&{mathbb P}left[frac{X}{X+Y} leq uright] ={mathbb P}left[Ygeq frac{(1-u)X}{u}right] \
=& 1-int_{0}^{infty}F_Yleft(frac{(1-u)x}{u}right)f_{X}(x) dx \
end{split}
end{equation}
Since you know both CDF and PDF of $chi^2$, you can evaluate $F_{U}(u)$ which may be with $_2 F_1(.,.,.,.)$ funtion. However, you may write $_2 F_1(.,.,.,.)$ with $B(.,.)$ function, if you really need it.
Then, you can find $f_{U}(u)=frac{d }{du}F_{U}(u)$.
answered May 3 '17 at 8:24
Frey
648312
answered May 3 '17 at 8:24
Frey
648312
answered May 3 '17 at 8:24
Frey
648312
answered May 3 '17 at 8:24
Frey
648312
648312
add a comment |
add a comment |
Let $X sim chi^2(m)$ and $Y sim chi^2(n)$, and let $T(X,Y)=(U,V)$ be a transformation such that $T_1(X,Y)=U=frac{X}{X+Y}$ and $T_2(X,Y)=V=X+Y$, then $T^{-1}_1(U,V)=X= UV$ and $T^{-1}_2(U,V)=Y=V-UV$. Then $ |J_{T^{-1}}|=|V| $. $X$ and $Y$ are independent then the joint density function is:
$$ f_{X,Y}(x,y)=frac{(1/2)^{m/2}}{Gamma(m/2)}x^{m/2 -1} e^{-x/2} frac{(1/2)^{n/2}}{Gamma(n/2)}y^{n/2 -1} e^{-y/2} $$
for $x, y geq 0$ and $0$ in other case.
Then, by the theorem of change of variable we have:
$$f_U(u)= int_{-infty}^{infty} frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} (uv)^{m/2-1}(v-uv)^{n/2-1}e^{-uv/2-(v-uv)/2} |v|dv $$
We have $v=x+y$ then $f_U$ is $0$ for $v < 0$ then we can consider the integral only in the interval $(0, infty)$ and $|v|=v$, also $u=frac{x}{x+y}$ then $0 leq u leq 1$. With some algebra:
$$f_U(u)= frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv $$
If $w=v/2$ and we make the change of variable in the integral we obtain:
$$int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv = 2^{m/2+n/2} Gamma(m/2+n/2)$$
Hence,
$$f_U(u)=frac{ Gamma(m/2+n/2) }{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} =dfrac{u^{m/2-1} (1-u)^{n/2-1}}{B(m/2,n/2)}$$
in the interval $[0,1]$, and this is the density function of the Beta distribution.
answered May 3 '17 at 10:26
Mike A.
567414
add a comment |
Let $X sim chi^2(m)$ and $Y sim chi^2(n)$, and let $T(X,Y)=(U,V)$ be a transformation such that $T_1(X,Y)=U=frac{X}{X+Y}$ and $T_2(X,Y)=V=X+Y$, then $T^{-1}_1(U,V)=X= UV$ and $T^{-1}_2(U,V)=Y=V-UV$. Then $ |J_{T^{-1}}|=|V| $. $X$ and $Y$ are independent then the joint density function is:
$$ f_{X,Y}(x,y)=frac{(1/2)^{m/2}}{Gamma(m/2)}x^{m/2 -1} e^{-x/2} frac{(1/2)^{n/2}}{Gamma(n/2)}y^{n/2 -1} e^{-y/2} $$
for $x, y geq 0$ and $0$ in other case.
Then, by the theorem of change of variable we have:
$$f_U(u)= int_{-infty}^{infty} frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} (uv)^{m/2-1}(v-uv)^{n/2-1}e^{-uv/2-(v-uv)/2} |v|dv $$
We have $v=x+y$ then $f_U$ is $0$ for $v < 0$ then we can consider the integral only in the interval $(0, infty)$ and $|v|=v$, also $u=frac{x}{x+y}$ then $0 leq u leq 1$. With some algebra:
$$f_U(u)= frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv $$
If $w=v/2$ and we make the change of variable in the integral we obtain:
$$int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv = 2^{m/2+n/2} Gamma(m/2+n/2)$$
Hence,
$$f_U(u)=frac{ Gamma(m/2+n/2) }{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} =dfrac{u^{m/2-1} (1-u)^{n/2-1}}{B(m/2,n/2)}$$
in the interval $[0,1]$, and this is the density function of the Beta distribution.
answered May 3 '17 at 10:26
Mike A.
567414
add a comment |
Let $X sim chi^2(m)$ and $Y sim chi^2(n)$, and let $T(X,Y)=(U,V)$ be a transformation such that $T_1(X,Y)=U=frac{X}{X+Y}$ and $T_2(X,Y)=V=X+Y$, then $T^{-1}_1(U,V)=X= UV$ and $T^{-1}_2(U,V)=Y=V-UV$. Then $ |J_{T^{-1}}|=|V| $. $X$ and $Y$ are independent then the joint density function is:
$$ f_{X,Y}(x,y)=frac{(1/2)^{m/2}}{Gamma(m/2)}x^{m/2 -1} e^{-x/2} frac{(1/2)^{n/2}}{Gamma(n/2)}y^{n/2 -1} e^{-y/2} $$
for $x, y geq 0$ and $0$ in other case.
Then, by the theorem of change of variable we have:
$$f_U(u)= int_{-infty}^{infty} frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} (uv)^{m/2-1}(v-uv)^{n/2-1}e^{-uv/2-(v-uv)/2} |v|dv $$
We have $v=x+y$ then $f_U$ is $0$ for $v < 0$ then we can consider the integral only in the interval $(0, infty)$ and $|v|=v$, also $u=frac{x}{x+y}$ then $0 leq u leq 1$. With some algebra:
$$f_U(u)= frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv $$
If $w=v/2$ and we make the change of variable in the integral we obtain:
$$int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv = 2^{m/2+n/2} Gamma(m/2+n/2)$$
Hence,
$$f_U(u)=frac{ Gamma(m/2+n/2) }{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} =dfrac{u^{m/2-1} (1-u)^{n/2-1}}{B(m/2,n/2)}$$
in the interval $[0,1]$, and this is the density function of the Beta distribution.
answered May 3 '17 at 10:26
Mike A.
567414
Let $X sim chi^2(m)$ and $Y sim chi^2(n)$, and let $T(X,Y)=(U,V)$ be a transformation such that $T_1(X,Y)=U=frac{X}{X+Y}$ and $T_2(X,Y)=V=X+Y$, then $T^{-1}_1(U,V)=X= UV$ and $T^{-1}_2(U,V)=Y=V-UV$. Then $ |J_{T^{-1}}|=|V| $. $X$ and $Y$ are independent then the joint density function is:
$$ f_{X,Y}(x,y)=frac{(1/2)^{m/2}}{Gamma(m/2)}x^{m/2 -1} e^{-x/2} frac{(1/2)^{n/2}}{Gamma(n/2)}y^{n/2 -1} e^{-y/2} $$
for $x, y geq 0$ and $0$ in other case.
Then, by the theorem of change of variable we have:
$$f_U(u)= int_{-infty}^{infty} frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} (uv)^{m/2-1}(v-uv)^{n/2-1}e^{-uv/2-(v-uv)/2} |v|dv $$
We have $v=x+y$ then $f_U$ is $0$ for $v < 0$ then we can consider the integral only in the interval $(0, infty)$ and $|v|=v$, also $u=frac{x}{x+y}$ then $0 leq u leq 1$. With some algebra:
$$f_U(u)= frac{(1/2)^{m/2+n/2}}{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv $$
If $w=v/2$ and we make the change of variable in the integral we obtain:
$$int_{0}^{infty} v^{m/2+n/2-1}e^{-v/2} dv = 2^{m/2+n/2} Gamma(m/2+n/2)$$
Hence,
$$f_U(u)=frac{ Gamma(m/2+n/2) }{Gamma(m/2) Gamma(n/2)} u^{m/2-1} (1-u)^{n/2-1} =dfrac{u^{m/2-1} (1-u)^{n/2-1}}{B(m/2,n/2)}$$
in the interval $[0,1]$, and this is the density function of the Beta distribution.
answered May 3 '17 at 10:26
Mike A.
567414
answered May 3 '17 at 10:26
Mike A.
567414
answered May 3 '17 at 10:26
Mike A.
567414
answered May 3 '17 at 10:26
Mike A.
567414
567414
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It is better to give the expression of $chi ^2 $, might help for the reader.
– Frey
May 3 '17 at 7:56
4
I'm pretty sure $X$ and $Z$ are not independent.
– Em.
May 3 '17 at 7:58
Thank you, the mistake was to consider $X$ and $Z$ independent
– Mike A.
May 3 '17 at 9:36