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Determining multivariate normality
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Let $X$ and $Y$ be i.i.d. $N(0, 1)$, and let $S$ be a random sign (1 or -1, with equal probabilities) independent of $(X, Y)$.
Determine whether or not $(SX, SY)$ is Multivariate Normal.
My attempt is:
$k_1SX + k_2SY = S(k_1X + k_2Y)$ for some constants $k_1$ and $k_2$
As $W = (k_1X + k_2Y)$ is normally distributed, we can show that $Z = SW$ is normally distributed as,
beginalign*
P(Z leq z) & = P(Zleq z|S=1)P(S=1) + P(Zleq z|S=-1)P(S=-1)\
& = 0.5P(W leq z)+0.5P(Wleq z)\
& = P(W leq z).
endalign*
Hence the distribution of $Z$ is normal for any choice of $k_1$ and $k_2$.
Is there anything wrong in this proof.
Attempt 2: As for the calculations given in Let X and Y be i.i.d. N(0, 1), and let S be a random sign (1 or -1, with equal probabilities) independent of (X, Y).
beginalign*
P((SX,SY)∈B)&=P((X,Y)∈B,S=1)+P((−X,−Y)∈B,S=−1) \
&= P((X,Y)∈B)P(S=1)+P((−X,−Y)∈B)P(S=−1)\
&= 0.5*P((X,Y)∈B) + 0.5*P((−X,−Y)∈B)\
&= P((X,Y)∈B)
endalign*
Which one is the correct answer?
probability-theory multivariable-calculus normal-distribution
asked Jul 21 at 11:27
Shana
408
add a comment |Â
up vote
1
down vote
favorite
Let $X$ and $Y$ be i.i.d. $N(0, 1)$, and let $S$ be a random sign (1 or -1, with equal probabilities) independent of $(X, Y)$.
Determine whether or not $(SX, SY)$ is Multivariate Normal.
My attempt is:
$k_1SX + k_2SY = S(k_1X + k_2Y)$ for some constants $k_1$ and $k_2$
As $W = (k_1X + k_2Y)$ is normally distributed, we can show that $Z = SW$ is normally distributed as,
beginalign*
P(Z leq z) & = P(Zleq z|S=1)P(S=1) + P(Zleq z|S=-1)P(S=-1)\
& = 0.5P(W leq z)+0.5P(Wleq z)\
& = P(W leq z).
endalign*
Hence the distribution of $Z$ is normal for any choice of $k_1$ and $k_2$.
Is there anything wrong in this proof.
Attempt 2: As for the calculations given in Let X and Y be i.i.d. N(0, 1), and let S be a random sign (1 or -1, with equal probabilities) independent of (X, Y).
beginalign*
P((SX,SY)∈B)&=P((X,Y)∈B,S=1)+P((−X,−Y)∈B,S=−1) \
&= P((X,Y)∈B)P(S=1)+P((−X,−Y)∈B)P(S=−1)\
&= 0.5*P((X,Y)∈B) + 0.5*P((−X,−Y)∈B)\
&= P((X,Y)∈B)
endalign*
Which one is the correct answer?
probability-theory multivariable-calculus normal-distribution
asked Jul 21 at 11:27
Shana
408
Same question: math.stackexchange.com/questions/1987113/…. See the comment.
– StubbornAtom
Jul 21 at 11:38
True. However, I am not sure how to solve it in that way. So I tried another way of getting it.
– Shana
Jul 21 at 11:44
You are using a theorem that two random variables have a joint normal distribution if all linear combinations are normal. The question is too elementary and using this theorem doesn't seem justified.
– Kavi Rama Murthy
Jul 21 at 12:01
If it is wrong, how can I prove or disprove multivariate normality in here?
– Shana
Jul 21 at 12:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ and $Y$ be i.i.d. $N(0, 1)$, and let $S$ be a random sign (1 or -1, with equal probabilities) independent of $(X, Y)$.
Determine whether or not $(SX, SY)$ is Multivariate Normal.
My attempt is:
$k_1SX + k_2SY = S(k_1X + k_2Y)$ for some constants $k_1$ and $k_2$
As $W = (k_1X + k_2Y)$ is normally distributed, we can show that $Z = SW$ is normally distributed as,
beginalign*
P(Z leq z) & = P(Zleq z|S=1)P(S=1) + P(Zleq z|S=-1)P(S=-1)\
& = 0.5P(W leq z)+0.5P(Wleq z)\
& = P(W leq z).
endalign*
Hence the distribution of $Z$ is normal for any choice of $k_1$ and $k_2$.
Is there anything wrong in this proof.
Attempt 2: As for the calculations given in Let X and Y be i.i.d. N(0, 1), and let S be a random sign (1 or -1, with equal probabilities) independent of (X, Y).
beginalign*
P((SX,SY)∈B)&=P((X,Y)∈B,S=1)+P((−X,−Y)∈B,S=−1) \
&= P((X,Y)∈B)P(S=1)+P((−X,−Y)∈B)P(S=−1)\
&= 0.5*P((X,Y)∈B) + 0.5*P((−X,−Y)∈B)\
&= P((X,Y)∈B)
endalign*
Which one is the correct answer?
probability-theory multivariable-calculus normal-distribution
asked Jul 21 at 11:27
Shana
408
Let $X$ and $Y$ be i.i.d. $N(0, 1)$, and let $S$ be a random sign (1 or -1, with equal probabilities) independent of $(X, Y)$.
Determine whether or not $(SX, SY)$ is Multivariate Normal.
My attempt is:
$k_1SX + k_2SY = S(k_1X + k_2Y)$ for some constants $k_1$ and $k_2$
As $W = (k_1X + k_2Y)$ is normally distributed, we can show that $Z = SW$ is normally distributed as,
beginalign*
P(Z leq z) & = P(Zleq z|S=1)P(S=1) + P(Zleq z|S=-1)P(S=-1)\
& = 0.5P(W leq z)+0.5P(Wleq z)\
& = P(W leq z).
endalign*
Hence the distribution of $Z$ is normal for any choice of $k_1$ and $k_2$.
Is there anything wrong in this proof.
Attempt 2: As for the calculations given in Let X and Y be i.i.d. N(0, 1), and let S be a random sign (1 or -1, with equal probabilities) independent of (X, Y).
beginalign*
P((SX,SY)∈B)&=P((X,Y)∈B,S=1)+P((−X,−Y)∈B,S=−1) \
&= P((X,Y)∈B)P(S=1)+P((−X,−Y)∈B)P(S=−1)\
&= 0.5*P((X,Y)∈B) + 0.5*P((−X,−Y)∈B)\
&= P((X,Y)∈B)
endalign*
Which one is the correct answer?
probability-theory multivariable-calculus normal-distribution
asked Jul 21 at 11:27
Shana
408
edited Jul 21 at 12:28
asked Jul 21 at 11:27
Shana
408
asked Jul 21 at 11:27
Shana
408
asked Jul 21 at 11:27
Shana
408
408
Same question: math.stackexchange.com/questions/1987113/…. See the comment.
– StubbornAtom
Jul 21 at 11:38
True. However, I am not sure how to solve it in that way. So I tried another way of getting it.
– Shana
Jul 21 at 11:44
You are using a theorem that two random variables have a joint normal distribution if all linear combinations are normal. The question is too elementary and using this theorem doesn't seem justified.
– Kavi Rama Murthy
Jul 21 at 12:01
If it is wrong, how can I prove or disprove multivariate normality in here?
– Shana
Jul 21 at 12:06
add a comment |Â
Same question: math.stackexchange.com/questions/1987113/…. See the comment.
– StubbornAtom
Jul 21 at 11:38
True. However, I am not sure how to solve it in that way. So I tried another way of getting it.
– Shana
Jul 21 at 11:44
You are using a theorem that two random variables have a joint normal distribution if all linear combinations are normal. The question is too elementary and using this theorem doesn't seem justified.
– Kavi Rama Murthy
Jul 21 at 12:01
If it is wrong, how can I prove or disprove multivariate normality in here?
– Shana
Jul 21 at 12:06
Same question: math.stackexchange.com/questions/1987113/…. See the comment.
– StubbornAtom
Jul 21 at 11:38
Same question: math.stackexchange.com/questions/1987113/…. See the comment.
– StubbornAtom
Jul 21 at 11:38
True. However, I am not sure how to solve it in that way. So I tried another way of getting it.
– Shana
Jul 21 at 11:44
True. However, I am not sure how to solve it in that way. So I tried another way of getting it.
– Shana
Jul 21 at 11:44
You are using a theorem that two random variables have a joint normal distribution if all linear combinations are normal. The question is too elementary and using this theorem doesn't seem justified.
– Kavi Rama Murthy
Jul 21 at 12:01
You are using a theorem that two random variables have a joint normal distribution if all linear combinations are normal. The question is too elementary and using this theorem doesn't seem justified.
– Kavi Rama Murthy
Jul 21 at 12:01
If it is wrong, how can I prove or disprove multivariate normality in here?
– Shana
Jul 21 at 12:06
If it is wrong, how can I prove or disprove multivariate normality in here?
– Shana
Jul 21 at 12:06
add a comment |Â
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Same question: math.stackexchange.com/questions/1987113/…. See the comment.
– StubbornAtom
Jul 21 at 11:38
True. However, I am not sure how to solve it in that way. So I tried another way of getting it.
– Shana
Jul 21 at 11:44
You are using a theorem that two random variables have a joint normal distribution if all linear combinations are normal. The question is too elementary and using this theorem doesn't seem justified.
– Kavi Rama Murthy
Jul 21 at 12:01
If it is wrong, how can I prove or disprove multivariate normality in here?
– Shana
Jul 21 at 12:06