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If joint probabilities equal mean their distributions are equal?
Clash Royale CLAN TAG#URR8PPP
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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 $(X, Y)$.
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*
How can we say that $(SX, SY)$ is multivariate normal?
probability-theory measure-theory probability-distributions
asked Jul 21 at 21:23
Shana
408
add a comment |Â
up vote
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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 $(X, Y)$.
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*
How can we say that $(SX, SY)$ is multivariate normal?
probability-theory measure-theory probability-distributions
asked Jul 21 at 21:23
Shana
408
The joint density function for $SX$ and $SY$ stays the same for each choice of $S$, so the joint density function is unchanged from the joint density of $X$ and $Y$.
– herb steinberg
Jul 21 at 21:41
@herbsteinberg If $P((SX, SY) in B) = P((X,Y)in B)$, does it mean joint pdf are necessarily equal?
– Shana
Jul 21 at 22:47
Yes, of course! By the very definition of joint distribution.
– Kavi Rama Murthy
Jul 21 at 23:37
@KaviRamaMurthy: I can see it intuitively. However, is it a general result or is it because the densities are continuous so that derivative of joint CDF gives joint PDF.
– Shana
Jul 21 at 23:56
The joint distribution of $(SX,SY)$ is that of $(X,Y)$, from which the result follows. That $X$ is a continuous random variable is irrelevant.
– Math1000
Jul 24 at 3:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
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 $(X, Y)$.
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*
How can we say that $(SX, SY)$ is multivariate normal?
probability-theory measure-theory probability-distributions
asked Jul 21 at 21:23
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 $(X, Y)$.
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*
How can we say that $(SX, SY)$ is multivariate normal?
probability-theory measure-theory probability-distributions
asked Jul 21 at 21:23
Shana
408
asked Jul 21 at 21:23
Shana
408
asked Jul 21 at 21:23
Shana
408
asked Jul 21 at 21:23
Shana
408
408
The joint density function for $SX$ and $SY$ stays the same for each choice of $S$, so the joint density function is unchanged from the joint density of $X$ and $Y$.
– herb steinberg
Jul 21 at 21:41
@herbsteinberg If $P((SX, SY) in B) = P((X,Y)in B)$, does it mean joint pdf are necessarily equal?
– Shana
Jul 21 at 22:47
Yes, of course! By the very definition of joint distribution.
– Kavi Rama Murthy
Jul 21 at 23:37
@KaviRamaMurthy: I can see it intuitively. However, is it a general result or is it because the densities are continuous so that derivative of joint CDF gives joint PDF.
– Shana
Jul 21 at 23:56
The joint distribution of $(SX,SY)$ is that of $(X,Y)$, from which the result follows. That $X$ is a continuous random variable is irrelevant.
– Math1000
Jul 24 at 3:11
add a comment |Â
The joint density function for $SX$ and $SY$ stays the same for each choice of $S$, so the joint density function is unchanged from the joint density of $X$ and $Y$.
– herb steinberg
Jul 21 at 21:41
@herbsteinberg If $P((SX, SY) in B) = P((X,Y)in B)$, does it mean joint pdf are necessarily equal?
– Shana
Jul 21 at 22:47
Yes, of course! By the very definition of joint distribution.
– Kavi Rama Murthy
Jul 21 at 23:37
@KaviRamaMurthy: I can see it intuitively. However, is it a general result or is it because the densities are continuous so that derivative of joint CDF gives joint PDF.
– Shana
Jul 21 at 23:56
The joint distribution of $(SX,SY)$ is that of $(X,Y)$, from which the result follows. That $X$ is a continuous random variable is irrelevant.
– Math1000
Jul 24 at 3:11
The joint density function for $SX$ and $SY$ stays the same for each choice of $S$, so the joint density function is unchanged from the joint density of $X$ and $Y$.
– herb steinberg
Jul 21 at 21:41
The joint density function for $SX$ and $SY$ stays the same for each choice of $S$, so the joint density function is unchanged from the joint density of $X$ and $Y$.
– herb steinberg
Jul 21 at 21:41
@herbsteinberg If $P((SX, SY) in B) = P((X,Y)in B)$, does it mean joint pdf are necessarily equal?
– Shana
Jul 21 at 22:47
@herbsteinberg If $P((SX, SY) in B) = P((X,Y)in B)$, does it mean joint pdf are necessarily equal?
– Shana
Jul 21 at 22:47
Yes, of course! By the very definition of joint distribution.
– Kavi Rama Murthy
Jul 21 at 23:37
Yes, of course! By the very definition of joint distribution.
– Kavi Rama Murthy
Jul 21 at 23:37
@KaviRamaMurthy: I can see it intuitively. However, is it a general result or is it because the densities are continuous so that derivative of joint CDF gives joint PDF.
– Shana
Jul 21 at 23:56
@KaviRamaMurthy: I can see it intuitively. However, is it a general result or is it because the densities are continuous so that derivative of joint CDF gives joint PDF.
– Shana
Jul 21 at 23:56
The joint distribution of $(SX,SY)$ is that of $(X,Y)$, from which the result follows. That $X$ is a continuous random variable is irrelevant.
– Math1000
Jul 24 at 3:11
The joint distribution of $(SX,SY)$ is that of $(X,Y)$, from which the result follows. That $X$ is a continuous random variable is irrelevant.
– Math1000
Jul 24 at 3:11
add a comment |Â
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The joint density function for $SX$ and $SY$ stays the same for each choice of $S$, so the joint density function is unchanged from the joint density of $X$ and $Y$.
– herb steinberg
Jul 21 at 21:41
@herbsteinberg If $P((SX, SY) in B) = P((X,Y)in B)$, does it mean joint pdf are necessarily equal?
– Shana
Jul 21 at 22:47
Yes, of course! By the very definition of joint distribution.
– Kavi Rama Murthy
Jul 21 at 23:37
@KaviRamaMurthy: I can see it intuitively. However, is it a general result or is it because the densities are continuous so that derivative of joint CDF gives joint PDF.
– Shana
Jul 21 at 23:56
The joint distribution of $(SX,SY)$ is that of $(X,Y)$, from which the result follows. That $X$ is a continuous random variable is irrelevant.
– Math1000
Jul 24 at 3:11