How to prove $P(X_1=x_1,…,X_n=x_n, X=X)=P(X_1=x_1)cdot …cdot P(X_n=x_n)$?
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Suppose $n>1$ and $X_1,X_2,...,X_n$ are $n$ independent random variables (say with expotentional distribution) with $X=sum_{i=1}^n X_n$.
Then $X_1,X_2,...,X_n,X$ are mutually independent random variables.
Thus according to the theorem (https://en.wikipedia.org/wiki/Joint_probability_distribution#Additional_properties) $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)cdot P(X=x)$.
However, the answer was $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)$.
How could that be?
probability statistics independence
asked 2 days ago
user9976437
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up vote
1
down vote
favorite
Suppose $n>1$ and $X_1,X_2,...,X_n$ are $n$ independent random variables (say with expotentional distribution) with $X=sum_{i=1}^n X_n$.
Then $X_1,X_2,...,X_n,X$ are mutually independent random variables.
Thus according to the theorem (https://en.wikipedia.org/wiki/Joint_probability_distribution#Additional_properties) $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)cdot P(X=x)$.
However, the answer was $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)$.
How could that be?
probability statistics independence
asked 2 days ago
user9976437
466
If $n=1$, it should be obvious that there is a dependence
– Hagen von Eitzen
2 days ago
$X$ is determined by the others. Also ${X=X}$ is an event with probability $1$, so the intersection of this event with any event $A$ has same the probability as $A$.
– Fnacool
2 days ago
@HagenvonEitzen But for $n>1$, $X$ and $X_i$ all have different probability distribution.
– user9976437
2 days ago
@Fnacool How to show they are dependent? And how to calculate this joint distribution? I'm not quite sure how to do this.
– user9976437
2 days ago
Do you know what dependent means?
– Mees de Vries
2 days ago
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $n>1$ and $X_1,X_2,...,X_n$ are $n$ independent random variables (say with expotentional distribution) with $X=sum_{i=1}^n X_n$.
Then $X_1,X_2,...,X_n,X$ are mutually independent random variables.
Thus according to the theorem (https://en.wikipedia.org/wiki/Joint_probability_distribution#Additional_properties) $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)cdot P(X=x)$.
However, the answer was $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)$.
How could that be?
probability statistics independence
asked 2 days ago
user9976437
466
Suppose $n>1$ and $X_1,X_2,...,X_n$ are $n$ independent random variables (say with expotentional distribution) with $X=sum_{i=1}^n X_n$.
Then $X_1,X_2,...,X_n,X$ are mutually independent random variables.
Thus according to the theorem (https://en.wikipedia.org/wiki/Joint_probability_distribution#Additional_properties) $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)cdot P(X=x)$.
However, the answer was $P(X_1=x_1,...,X_n=x_n, X=X)=P(X_1=x_1)cdot ...cdot P(X_n=x_n)$.
How could that be?
probability statistics independence
probability statistics independence
asked 2 days ago
user9976437
466
asked 2 days ago
user9976437
466
edited 2 days ago
asked 2 days ago
user9976437
466
asked 2 days ago
user9976437
466
asked 2 days ago
user9976437
466
466
If $n=1$, it should be obvious that there is a dependence
– Hagen von Eitzen
2 days ago
$X$ is determined by the others. Also ${X=X}$ is an event with probability $1$, so the intersection of this event with any event $A$ has same the probability as $A$.
– Fnacool
2 days ago
@HagenvonEitzen But for $n>1$, $X$ and $X_i$ all have different probability distribution.
– user9976437
2 days ago
@Fnacool How to show they are dependent? And how to calculate this joint distribution? I'm not quite sure how to do this.
– user9976437
2 days ago
Do you know what dependent means?
– Mees de Vries
2 days ago
|
show 1 more comment
If $n=1$, it should be obvious that there is a dependence
– Hagen von Eitzen
2 days ago
$X$ is determined by the others. Also ${X=X}$ is an event with probability $1$, so the intersection of this event with any event $A$ has same the probability as $A$.
– Fnacool
2 days ago
@HagenvonEitzen But for $n>1$, $X$ and $X_i$ all have different probability distribution.
– user9976437
2 days ago
@Fnacool How to show they are dependent? And how to calculate this joint distribution? I'm not quite sure how to do this.
– user9976437
2 days ago
Do you know what dependent means?
– Mees de Vries
2 days ago
If $n=1$, it should be obvious that there is a dependence
– Hagen von Eitzen
2 days ago
If $n=1$, it should be obvious that there is a dependence
– Hagen von Eitzen
2 days ago
$X$ is determined by the others. Also ${X=X}$ is an event with probability $1$, so the intersection of this event with any event $A$ has same the probability as $A$.
– Fnacool
2 days ago
$X$ is determined by the others. Also ${X=X}$ is an event with probability $1$, so the intersection of this event with any event $A$ has same the probability as $A$.
– Fnacool
2 days ago
@HagenvonEitzen But for $n>1$, $X$ and $X_i$ all have different probability distribution.
– user9976437
2 days ago
@HagenvonEitzen But for $n>1$, $X$ and $X_i$ all have different probability distribution.
– user9976437
2 days ago
@Fnacool How to show they are dependent? And how to calculate this joint distribution? I'm not quite sure how to do this.
– user9976437
2 days ago
@Fnacool How to show they are dependent? And how to calculate this joint distribution? I'm not quite sure how to do this.
– user9976437
2 days ago
Do you know what dependent means?
– Mees de Vries
2 days ago
Do you know what dependent means?
– Mees de Vries
2 days ago
|
show 1 more comment
1 Answer
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oldest
votes
up vote
0
down vote
Random variable X is dependent on the other random variables.
$because$ Take an example, Suppose $X_i = i forall iin{1,2,...,n}$.Then we can completely tell whether for a particular n-dimensional vector $x, X=x$ or not.
$therefore$ The random variable X is not independent to other random variables,however, the result is true for $X_i's$ separately,i.e. $$P(X_1=x_1,X_2=x_2,...,X_n=x_n) = prod_{i=1}^nP(X_i=x_i)$$
Hope it helps:)
answered 2 days ago
Crazy for maths
4736
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Random variable X is dependent on the other random variables.
$because$ Take an example, Suppose $X_i = i forall iin{1,2,...,n}$.Then we can completely tell whether for a particular n-dimensional vector $x, X=x$ or not.
$therefore$ The random variable X is not independent to other random variables,however, the result is true for $X_i's$ separately,i.e. $$P(X_1=x_1,X_2=x_2,...,X_n=x_n) = prod_{i=1}^nP(X_i=x_i)$$
Hope it helps:)
answered 2 days ago
Crazy for maths
4736
add a comment |
up vote
0
down vote
Random variable X is dependent on the other random variables.
$because$ Take an example, Suppose $X_i = i forall iin{1,2,...,n}$.Then we can completely tell whether for a particular n-dimensional vector $x, X=x$ or not.
$therefore$ The random variable X is not independent to other random variables,however, the result is true for $X_i's$ separately,i.e. $$P(X_1=x_1,X_2=x_2,...,X_n=x_n) = prod_{i=1}^nP(X_i=x_i)$$
Hope it helps:)
answered 2 days ago
Crazy for maths
4736
add a comment |
up vote
0
down vote
up vote
0
down vote
Random variable X is dependent on the other random variables.
$because$ Take an example, Suppose $X_i = i forall iin{1,2,...,n}$.Then we can completely tell whether for a particular n-dimensional vector $x, X=x$ or not.
$therefore$ The random variable X is not independent to other random variables,however, the result is true for $X_i's$ separately,i.e. $$P(X_1=x_1,X_2=x_2,...,X_n=x_n) = prod_{i=1}^nP(X_i=x_i)$$
Hope it helps:)
answered 2 days ago
Crazy for maths
4736
Random variable X is dependent on the other random variables.
$because$ Take an example, Suppose $X_i = i forall iin{1,2,...,n}$.Then we can completely tell whether for a particular n-dimensional vector $x, X=x$ or not.
$therefore$ The random variable X is not independent to other random variables,however, the result is true for $X_i's$ separately,i.e. $$P(X_1=x_1,X_2=x_2,...,X_n=x_n) = prod_{i=1}^nP(X_i=x_i)$$
Hope it helps:)
answered 2 days ago
Crazy for maths
4736
answered 2 days ago
Crazy for maths
4736
answered 2 days ago
Crazy for maths
4736
answered 2 days ago
Crazy for maths
4736
4736
add a comment |
add a comment |
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If $n=1$, it should be obvious that there is a dependence
– Hagen von Eitzen
2 days ago
$X$ is determined by the others. Also ${X=X}$ is an event with probability $1$, so the intersection of this event with any event $A$ has same the probability as $A$.
– Fnacool
2 days ago
@HagenvonEitzen But for $n>1$, $X$ and $X_i$ all have different probability distribution.
– user9976437
2 days ago
@Fnacool How to show they are dependent? And how to calculate this joint distribution? I'm not quite sure how to do this.
– user9976437
2 days ago
Do you know what dependent means?
– Mees de Vries
2 days ago