What functions can be used to dilate a probability distribution function?
$begingroup$
Convolving any distribution $p(x)$ by a $delta(t-tau)$ will horizontally shift the distribution by $tau$. What mathematical operation can I perform on $p(x)$ that will horizontally dilate it (i.e., I want to generate some function $g(x)$ where $g(3x)=p(x)$). In my specific case, I'm working with probability distribution functions that I want to horizontally dilate by some constant.
New here, so please let me know if I should've asked the above question in a different way! Thanks!
Edit:
I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
geometry statistics functions
asked Dec 18 '18 at 2:19
regenerationregeneration
11
$endgroup$
add a comment |
$begingroup$
Convolving any distribution $p(x)$ by a $delta(t-tau)$ will horizontally shift the distribution by $tau$. What mathematical operation can I perform on $p(x)$ that will horizontally dilate it (i.e., I want to generate some function $g(x)$ where $g(3x)=p(x)$). In my specific case, I'm working with probability distribution functions that I want to horizontally dilate by some constant.
New here, so please let me know if I should've asked the above question in a different way! Thanks!
Edit:
I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
geometry statistics functions
asked Dec 18 '18 at 2:19
regenerationregeneration
11
$endgroup$
1
$begingroup$
Is $p(x)$ a probability density or a cumulative distribution function?
$endgroup$
– parsiad
Dec 18 '18 at 2:24
add a comment |
$begingroup$
Convolving any distribution $p(x)$ by a $delta(t-tau)$ will horizontally shift the distribution by $tau$. What mathematical operation can I perform on $p(x)$ that will horizontally dilate it (i.e., I want to generate some function $g(x)$ where $g(3x)=p(x)$). In my specific case, I'm working with probability distribution functions that I want to horizontally dilate by some constant.
New here, so please let me know if I should've asked the above question in a different way! Thanks!
Edit:
I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
geometry statistics functions
asked Dec 18 '18 at 2:19
regenerationregeneration
11
$endgroup$
Convolving any distribution $p(x)$ by a $delta(t-tau)$ will horizontally shift the distribution by $tau$. What mathematical operation can I perform on $p(x)$ that will horizontally dilate it (i.e., I want to generate some function $g(x)$ where $g(3x)=p(x)$). In my specific case, I'm working with probability distribution functions that I want to horizontally dilate by some constant.
New here, so please let me know if I should've asked the above question in a different way! Thanks!
Edit:
I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
geometry statistics functions
geometry statistics functions
asked Dec 18 '18 at 2:19
regenerationregeneration
11
asked Dec 18 '18 at 2:19
regenerationregeneration
11
edited Dec 18 '18 at 4:31
regeneration
asked Dec 18 '18 at 2:19
regenerationregeneration
11
asked Dec 18 '18 at 2:19
regenerationregeneration
11
asked Dec 18 '18 at 2:19
regenerationregeneration
11
11
1
$begingroup$
Is $p(x)$ a probability density or a cumulative distribution function?
$endgroup$
– parsiad
Dec 18 '18 at 2:24
add a comment |
1
$begingroup$
Is $p(x)$ a probability density or a cumulative distribution function?
$endgroup$
– parsiad
Dec 18 '18 at 2:24
1
1
$begingroup$
Is $p(x)$ a probability density or a cumulative distribution function?
$endgroup$
– parsiad
Dec 18 '18 at 2:24
$begingroup$
Is $p(x)$ a probability density or a cumulative distribution function?
$endgroup$
– parsiad
Dec 18 '18 at 2:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm assuming you meant probability density.
You know that
$$
int_{-infty}^{infty}p(x)dx=1.
$$
Using the substitution rule with $x=3y$,
$$
int_{-infty}^{infty}3p(3y)dy=1.
$$
So your "dilated" probability density is $ymapsto3p(3y)$.
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
$endgroup$
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
I'm assuming you meant probability density.
You know that
$$
int_{-infty}^{infty}p(x)dx=1.
$$
Using the substitution rule with $x=3y$,
$$
int_{-infty}^{infty}3p(3y)dy=1.
$$
So your "dilated" probability density is $ymapsto3p(3y)$.
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
$endgroup$
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
add a comment |
$begingroup$
I'm assuming you meant probability density.
You know that
$$
int_{-infty}^{infty}p(x)dx=1.
$$
Using the substitution rule with $x=3y$,
$$
int_{-infty}^{infty}3p(3y)dy=1.
$$
So your "dilated" probability density is $ymapsto3p(3y)$.
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
$endgroup$
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
add a comment |
$begingroup$
I'm assuming you meant probability density.
You know that
$$
int_{-infty}^{infty}p(x)dx=1.
$$
Using the substitution rule with $x=3y$,
$$
int_{-infty}^{infty}3p(3y)dy=1.
$$
So your "dilated" probability density is $ymapsto3p(3y)$.
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
$endgroup$
I'm assuming you meant probability density.
You know that
$$
int_{-infty}^{infty}p(x)dx=1.
$$
Using the substitution rule with $x=3y$,
$$
int_{-infty}^{infty}3p(3y)dy=1.
$$
So your "dilated" probability density is $ymapsto3p(3y)$.
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
answered Dec 18 '18 at 2:32
parsiadparsiad
18.3k32453
18.3k32453
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
add a comment |
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Yes, I mean probability density. To be more clear, I'm looking for a function that can be applied to p(x) directly instead of through substitution (i.e., $p(x) ?h(x)=g(x))$, where ? is some operator and $h(x)$ is some function/distribution. In the shifting example above, the delta function would be h(x).
$endgroup$
– regeneration
Dec 18 '18 at 4:31
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
Technically, $G$ defined by $(Gp)(y) = 3p(3y)$ is an operator... ;-)
$endgroup$
– parsiad
Dec 18 '18 at 4:33
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
You should narrow down your problem though, as it stands, it's too broad. Maybe pick a specific operator (e.g., multiplication).
$endgroup$
– parsiad
Dec 18 '18 at 4:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
$begingroup$
Hmm I see. I've been thinking about mostly convolution or multiplication. In your example above you still don't derive an h(x) for $p(x)?h(x)=g(x)$. I'd like a form where I have both an operator (multiplication or convolution, ideally) and a function that does the transformation h(x).
$endgroup$
– regeneration
Dec 18 '18 at 16:35
add a comment |
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$begingroup$
Is $p(x)$ a probability density or a cumulative distribution function?
$endgroup$
– parsiad
Dec 18 '18 at 2:24