Mixing
Find the pdf of y=x^2+1
$begingroup$
I have a random variable X with pdf
(-1 0 1 2)
(1/4 1/2 1/8 1/8)
and I need to find the pdf of Y = X^2 + 1.
So by squaring X and adding 1 I obtain values 1, 2, 5 for the X vector. But how should I calculate the corresponding probabilities?
(1 2 5)
(? ? ?)
Thank you.
probability-theory
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
$endgroup$
add a comment |
$begingroup$
I have a random variable X with pdf
(-1 0 1 2)
(1/4 1/2 1/8 1/8)
and I need to find the pdf of Y = X^2 + 1.
So by squaring X and adding 1 I obtain values 1, 2, 5 for the X vector. But how should I calculate the corresponding probabilities?
(1 2 5)
(? ? ?)
Thank you.
probability-theory
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
$endgroup$
$begingroup$
For each possible value of $X^2+1$ you need to find the corresponding value(s) of $X$ and their probabilities
$endgroup$
– Henry
Feb 3 at 12:04
add a comment |
$begingroup$
I have a random variable X with pdf
(-1 0 1 2)
(1/4 1/2 1/8 1/8)
and I need to find the pdf of Y = X^2 + 1.
So by squaring X and adding 1 I obtain values 1, 2, 5 for the X vector. But how should I calculate the corresponding probabilities?
(1 2 5)
(? ? ?)
Thank you.
probability-theory
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
$endgroup$
I have a random variable X with pdf
(-1 0 1 2)
(1/4 1/2 1/8 1/8)
and I need to find the pdf of Y = X^2 + 1.
So by squaring X and adding 1 I obtain values 1, 2, 5 for the X vector. But how should I calculate the corresponding probabilities?
(1 2 5)
(? ? ?)
Thank you.
probability-theory
probability-theory
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
asked Feb 3 at 12:00
Razvan AxinieRazvan Axinie
132
132
$begingroup$
For each possible value of $X^2+1$ you need to find the corresponding value(s) of $X$ and their probabilities
$endgroup$
– Henry
Feb 3 at 12:04
add a comment |
$begingroup$
For each possible value of $X^2+1$ you need to find the corresponding value(s) of $X$ and their probabilities
$endgroup$
– Henry
Feb 3 at 12:04
$begingroup$
For each possible value of $X^2+1$ you need to find the corresponding value(s) of $X$ and their probabilities
$endgroup$
– Henry
Feb 3 at 12:04
$begingroup$
For each possible value of $X^2+1$ you need to find the corresponding value(s) of $X$ and their probabilities
$endgroup$
– Henry
Feb 3 at 12:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We simply add the associated probabilities of $X$ for each of the given values which you squared and added $1$ to.
$$(1,2,5) Rightarrow (frac{1}{2},frac{1}{4}+frac{1}{8},frac{1}{8})$$
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
$endgroup$
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
We simply add the associated probabilities of $X$ for each of the given values which you squared and added $1$ to.
$$(1,2,5) Rightarrow (frac{1}{2},frac{1}{4}+frac{1}{8},frac{1}{8})$$
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
$endgroup$
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
add a comment |
$begingroup$
We simply add the associated probabilities of $X$ for each of the given values which you squared and added $1$ to.
$$(1,2,5) Rightarrow (frac{1}{2},frac{1}{4}+frac{1}{8},frac{1}{8})$$
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
$endgroup$
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
add a comment |
$begingroup$
We simply add the associated probabilities of $X$ for each of the given values which you squared and added $1$ to.
$$(1,2,5) Rightarrow (frac{1}{2},frac{1}{4}+frac{1}{8},frac{1}{8})$$
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
$endgroup$
We simply add the associated probabilities of $X$ for each of the given values which you squared and added $1$ to.
$$(1,2,5) Rightarrow (frac{1}{2},frac{1}{4}+frac{1}{8},frac{1}{8})$$
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
answered Feb 3 at 12:04
Peter ForemanPeter Foreman
7,5361320
7,5361320
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
add a comment |
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
So it's like having a mapping (old value) -> (new value) for every vector component and the probabilities stay the same, and when we have overlapping (new values) we add the probabilities from the corresponding (old values) ?
$endgroup$
– Razvan Axinie
Feb 3 at 12:11
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
$begingroup$
Yes exactly like that. This is because $P(f(X)=f(x))=P(X=x)$.
$endgroup$
– Peter Foreman
Feb 3 at 12:15
add a comment |
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$begingroup$
For each possible value of $X^2+1$ you need to find the corresponding value(s) of $X$ and their probabilities
$endgroup$
– Henry
Feb 3 at 12:04