Mixing
Given the cumulative distribution function find a random variable that has this distribution.
$begingroup$
We are given a Cumulative distribution function $(CDF)$ how do we find a random variable that has the given distribution? Since there could be a lot of such variables we are to find anyone given that the function is differentiable and increasing.
I read somewhere that we could simply take inverse of the function(if it exists) and that would be the Random variable you are looking for, but I don't understand it, even if it's true(?).
If that's true, why? Else, how do we find a random variable for the distribution?
probability statistics probability-distributions random-variables
asked Aug 8 '17 at 12:38
aromaaroma
217211
$endgroup$
|
show 4 more comments
$begingroup$
We are given a Cumulative distribution function $(CDF)$ how do we find a random variable that has the given distribution? Since there could be a lot of such variables we are to find anyone given that the function is differentiable and increasing.
I read somewhere that we could simply take inverse of the function(if it exists) and that would be the Random variable you are looking for, but I don't understand it, even if it's true(?).
If that's true, why? Else, how do we find a random variable for the distribution?
probability statistics probability-distributions random-variables
asked Aug 8 '17 at 12:38
aromaaroma
217211
$endgroup$
$begingroup$
I see that the answers discuss prob integral transform (en.wikipedia.org/wiki/Probability_integral_transform), which answers/clarifies the second paragraph of the question of generating a rv $X$ that has the given CDF from a standard uniform rv. But I don't know what the question in the titlegiven the cumulative distribution function find a random variable that has this distributionmeans since if a cdf $F(x)$ is given for some random variable $X$ then for any $a$ and $b$ the probability $P(a < X leq b)$ is nailed down by $F(b)-F(a)$ so $F(x)$ does indeed specify the rv a.e.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 16:16
$begingroup$
@Just_to_Answer Obviously F(x) is a random variable, but it's distribution isn't the same as F(x).
$endgroup$
– aroma
Aug 8 '17 at 17:31
$begingroup$
$F(x)$ by convention is not a random variable. It is a real valued function $mathbb R rightarrow [0,1]$ defined by $F(x) = P(X leq x)$. So my point is specifying $F(x)$ is a.e. same as specifying $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:41
$begingroup$
@Just_to_Answer I don't understand your point, even if $F(x)$ is a r.v. how does it help in finding a r.v. whose CDF is $F(x)$. Since CDF of the r.v. $F(x)$ is not necessarily $F(x)$.
$endgroup$
– aroma
Aug 8 '17 at 17:45
$begingroup$
Let me try it differently. You give me a valid CDF $F(x)$ and say this is the CDF of some rv $X$. Using just that $F(x)$ and nothing else, I can compute any probabilities, moments, etc.etc. involving the rv $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:52
|
show 4 more comments
$begingroup$
We are given a Cumulative distribution function $(CDF)$ how do we find a random variable that has the given distribution? Since there could be a lot of such variables we are to find anyone given that the function is differentiable and increasing.
I read somewhere that we could simply take inverse of the function(if it exists) and that would be the Random variable you are looking for, but I don't understand it, even if it's true(?).
If that's true, why? Else, how do we find a random variable for the distribution?
probability statistics probability-distributions random-variables
asked Aug 8 '17 at 12:38
aromaaroma
217211
$endgroup$
We are given a Cumulative distribution function $(CDF)$ how do we find a random variable that has the given distribution? Since there could be a lot of such variables we are to find anyone given that the function is differentiable and increasing.
I read somewhere that we could simply take inverse of the function(if it exists) and that would be the Random variable you are looking for, but I don't understand it, even if it's true(?).
If that's true, why? Else, how do we find a random variable for the distribution?
probability statistics probability-distributions random-variables
probability statistics probability-distributions random-variables
asked Aug 8 '17 at 12:38
aromaaroma
217211
asked Aug 8 '17 at 12:38
aromaaroma
217211
asked Aug 8 '17 at 12:38
aromaaroma
217211
asked Aug 8 '17 at 12:38
aromaaroma
217211
asked Aug 8 '17 at 12:38
aromaaroma
217211
217211
$begingroup$
I see that the answers discuss prob integral transform (en.wikipedia.org/wiki/Probability_integral_transform), which answers/clarifies the second paragraph of the question of generating a rv $X$ that has the given CDF from a standard uniform rv. But I don't know what the question in the titlegiven the cumulative distribution function find a random variable that has this distributionmeans since if a cdf $F(x)$ is given for some random variable $X$ then for any $a$ and $b$ the probability $P(a < X leq b)$ is nailed down by $F(b)-F(a)$ so $F(x)$ does indeed specify the rv a.e.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 16:16
$begingroup$
@Just_to_Answer Obviously F(x) is a random variable, but it's distribution isn't the same as F(x).
$endgroup$
– aroma
Aug 8 '17 at 17:31
$begingroup$
$F(x)$ by convention is not a random variable. It is a real valued function $mathbb R rightarrow [0,1]$ defined by $F(x) = P(X leq x)$. So my point is specifying $F(x)$ is a.e. same as specifying $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:41
$begingroup$
@Just_to_Answer I don't understand your point, even if $F(x)$ is a r.v. how does it help in finding a r.v. whose CDF is $F(x)$. Since CDF of the r.v. $F(x)$ is not necessarily $F(x)$.
$endgroup$
– aroma
Aug 8 '17 at 17:45
$begingroup$
Let me try it differently. You give me a valid CDF $F(x)$ and say this is the CDF of some rv $X$. Using just that $F(x)$ and nothing else, I can compute any probabilities, moments, etc.etc. involving the rv $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:52
|
show 4 more comments
$begingroup$
I see that the answers discuss prob integral transform (en.wikipedia.org/wiki/Probability_integral_transform), which answers/clarifies the second paragraph of the question of generating a rv $X$ that has the given CDF from a standard uniform rv. But I don't know what the question in the titlegiven the cumulative distribution function find a random variable that has this distributionmeans since if a cdf $F(x)$ is given for some random variable $X$ then for any $a$ and $b$ the probability $P(a < X leq b)$ is nailed down by $F(b)-F(a)$ so $F(x)$ does indeed specify the rv a.e.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 16:16
$begingroup$
@Just_to_Answer Obviously F(x) is a random variable, but it's distribution isn't the same as F(x).
$endgroup$
– aroma
Aug 8 '17 at 17:31
$begingroup$
$F(x)$ by convention is not a random variable. It is a real valued function $mathbb R rightarrow [0,1]$ defined by $F(x) = P(X leq x)$. So my point is specifying $F(x)$ is a.e. same as specifying $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:41
$begingroup$
@Just_to_Answer I don't understand your point, even if $F(x)$ is a r.v. how does it help in finding a r.v. whose CDF is $F(x)$. Since CDF of the r.v. $F(x)$ is not necessarily $F(x)$.
$endgroup$
– aroma
Aug 8 '17 at 17:45
$begingroup$
Let me try it differently. You give me a valid CDF $F(x)$ and say this is the CDF of some rv $X$. Using just that $F(x)$ and nothing else, I can compute any probabilities, moments, etc.etc. involving the rv $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:52
$begingroup$
I see that the answers discuss prob integral transform (en.wikipedia.org/wiki/Probability_integral_transform), which answers/clarifies the second paragraph of the question of generating a rv $X$ that has the given CDF from a standard uniform rv. But I don't know what the question in the title
given the cumulative distribution function find a random variable that has this distribution means since if a cdf $F(x)$ is given for some random variable $X$ then for any $a$ and $b$ the probability $P(a < X leq b)$ is nailed down by $F(b)-F(a)$ so $F(x)$ does indeed specify the rv a.e.$endgroup$
– Just_to_Answer
Aug 8 '17 at 16:16
$begingroup$
I see that the answers discuss prob integral transform (en.wikipedia.org/wiki/Probability_integral_transform), which answers/clarifies the second paragraph of the question of generating a rv $X$ that has the given CDF from a standard uniform rv. But I don't know what the question in the title
given the cumulative distribution function find a random variable that has this distribution means since if a cdf $F(x)$ is given for some random variable $X$ then for any $a$ and $b$ the probability $P(a < X leq b)$ is nailed down by $F(b)-F(a)$ so $F(x)$ does indeed specify the rv a.e.$endgroup$
– Just_to_Answer
Aug 8 '17 at 16:16
$begingroup$
@Just_to_Answer Obviously F(x) is a random variable, but it's distribution isn't the same as F(x).
$endgroup$
– aroma
Aug 8 '17 at 17:31
$begingroup$
@Just_to_Answer Obviously F(x) is a random variable, but it's distribution isn't the same as F(x).
$endgroup$
– aroma
Aug 8 '17 at 17:31
$begingroup$
$F(x)$ by convention is not a random variable. It is a real valued function $mathbb R rightarrow [0,1]$ defined by $F(x) = P(X leq x)$. So my point is specifying $F(x)$ is a.e. same as specifying $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:41
$begingroup$
$F(x)$ by convention is not a random variable. It is a real valued function $mathbb R rightarrow [0,1]$ defined by $F(x) = P(X leq x)$. So my point is specifying $F(x)$ is a.e. same as specifying $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:41
$begingroup$
@Just_to_Answer I don't understand your point, even if $F(x)$ is a r.v. how does it help in finding a r.v. whose CDF is $F(x)$. Since CDF of the r.v. $F(x)$ is not necessarily $F(x)$.
$endgroup$
– aroma
Aug 8 '17 at 17:45
$begingroup$
@Just_to_Answer I don't understand your point, even if $F(x)$ is a r.v. how does it help in finding a r.v. whose CDF is $F(x)$. Since CDF of the r.v. $F(x)$ is not necessarily $F(x)$.
$endgroup$
– aroma
Aug 8 '17 at 17:45
$begingroup$
Let me try it differently. You give me a valid CDF $F(x)$ and say this is the CDF of some rv $X$. Using just that $F(x)$ and nothing else, I can compute any probabilities, moments, etc.etc. involving the rv $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:52
$begingroup$
Let me try it differently. You give me a valid CDF $F(x)$ and say this is the CDF of some rv $X$. Using just that $F(x)$ and nothing else, I can compute any probabilities, moments, etc.etc. involving the rv $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:52
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Let $X=g^{-1}(Y)$ where $Y$ is uniform on $[0,1]$ and $g(cdot)$ is strictly increasing (and hence invertible).
Then $$F_X(x)=P(Xle x)=P(g^{-1}(Y) le x) = P(Y le g(x))=g(x) $$
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
$endgroup$
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
add a comment |
$begingroup$
Let $U$ be a random variable with uniform distribution on $(0,1)$.
Prescribe $Phi:(0,1)tomathbb R$ by:$$umapstoinf{xinmathbb Rmid uleq F(x)}$$
Then it can be shown that:$$Phi(u)leq xiff uleq F(x)$$
So letting $U$ be a random variable with uniform distribution on $(0,1)$ we have:$$Phi(U)leq xiff Uleq F(x)$$and consequently:$$F_{Phi(U)}(x)=P(Phi(u)leq x)=P(Uleq F(x))=F(x)$$
This always works.
In the special case where $F$ has an inverse we find that $Phi$ is actually that inverse.
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
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add a comment |
$begingroup$
If you are studying elementary probability theory, allow me to reformulate your question as "how can I represent a random variable $X$ with a given CDF $F_X$ in terms of a uniform random variable $U$ on $(0,1)$?" The answer to that is the quantile function: you define
$$G_X(p)=inf { x : F_X(x) geq p }$$
and then define $X$ to be $G_X(U)$.
Note that if $F_X$ is invertible then $G_X=F_X^{-1}$, otherwise this is "the right generalization". One can see this by looking at the discrete case: if $P(X=x)=p$ then $P(G_X(U)=x)=p$. This is because a jump of height $p$ in $F_X$ corresponds to a flat region of length $p$ in $G_X$, and the uniform distribution on $(0,1)$ assigns each interval a probability equal to its length.
The natural question is now "what's a uniform random variable on $(0,1)$?" Well, it has $F_U(x)=begin{cases} 0 & x<0 \ x & x in [0,1] \ 1 & x>1 end{cases}$. But otherwise such a thing is a black box from the elementary point of view.
If you are studying measure-theoretic probability theory then the answer is a bit more explicit. A random variable with CDF $F_X$ is given by $G_X : Omega to mathbb{R}$ where $G_X$ is the quantile function as defined before, $Omega=(0,1)$, $mathcal{F}$ is the Borel $sigma$-algebra on $(0,1)$, and $mathbb{P}$ is the Lebesgue measure. Note that on this space the identity function is a uniform random variable on $(0,1)$, so this is really the same construction as the one described above.
In any case these constructions can be generalized to finitely many random variables by looking at the uniform distribution on $(0,1)^n$ instead of $(0,1)$.
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
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1
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Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
add a comment |
$begingroup$
The $text{cdf}$ of a random variable tells the probability that its value falls below a given bound,
$$text{cdf}_X(x)=mathbb P(X<x).$$
For a uniform variable $U$, we have
$$text{cdf}_U(x)=mathbb P(U<x)=x$$ for $xin[0,1]$.
Now if you want to create a random variable with an imposed $text{cdf}$, let $h(x)$, you want to achieve
$$mathbb P(X<x)=h(x).$$
If $X$ derives from a uniform variable via some transformation, let $X=g(U)$, you have
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x).$$
But also, assuming $g$ invertible, and using the fact that $U$ is uniform,
$$mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=g^{-1}(x).$$
Putting these facts together,
$$g^{-1}(x)=h(x)$$ or $$g(x)=h^{-1}(x).$$
For a simple example, take a negative exponential distribution,
$$text{cdf}_X=h(x)=1-e^{-x}.$$
Then by inversion,
$$g(x)=-ln(1-x)$$ which maps $0$ to $0$ and $1$ to $infty$. By looking at the plot of $g$, you will understand that the "density" of the transformed points goes decreasing, because the slope increases, resulting in a bias in favor of the small values.
We can even play the exercise when $U$ is not uniform but has a known $text{cdf}$, let $f(x)$. Then it suffices to adapt the equations, and
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x)=mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=f(g^{-1})(x)$$ and we need
$$g(x)=h^{-1}(f(x)).$$
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
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1
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The right-continuous convention for CDFs is much more common these days than the left-continuous one.
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– Ian
Aug 8 '17 at 13:19
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@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
1
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Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
add a comment |
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4 Answers
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4 Answers
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$begingroup$
Let $X=g^{-1}(Y)$ where $Y$ is uniform on $[0,1]$ and $g(cdot)$ is strictly increasing (and hence invertible).
Then $$F_X(x)=P(Xle x)=P(g^{-1}(Y) le x) = P(Y le g(x))=g(x) $$
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
$endgroup$
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
add a comment |
$begingroup$
Let $X=g^{-1}(Y)$ where $Y$ is uniform on $[0,1]$ and $g(cdot)$ is strictly increasing (and hence invertible).
Then $$F_X(x)=P(Xle x)=P(g^{-1}(Y) le x) = P(Y le g(x))=g(x) $$
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
$endgroup$
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
add a comment |
$begingroup$
Let $X=g^{-1}(Y)$ where $Y$ is uniform on $[0,1]$ and $g(cdot)$ is strictly increasing (and hence invertible).
Then $$F_X(x)=P(Xle x)=P(g^{-1}(Y) le x) = P(Y le g(x))=g(x) $$
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
$endgroup$
Let $X=g^{-1}(Y)$ where $Y$ is uniform on $[0,1]$ and $g(cdot)$ is strictly increasing (and hence invertible).
Then $$F_X(x)=P(Xle x)=P(g^{-1}(Y) le x) = P(Y le g(x))=g(x) $$
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
answered Aug 8 '17 at 12:58
leonbloyleonbloy
41.6k647108
41.6k647108
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
add a comment |
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
This looks mathematically fine, but is there way to intuitively understand this?
$endgroup$
– aroma
Aug 8 '17 at 13:01
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
Perhaps it's a little easier to think it the other way math.stackexchange.com/questions/868400/…
$endgroup$
– leonbloy
Aug 8 '17 at 13:06
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
$begingroup$
One intuitive way to think about the concept above (inverse transform sampling) is this: Visualize the graph of your CDF which has the range $(0,1)$. Imagine picking a point randomly in the range $(0,1)$ on the vertical axis, then going horizontally until you hit the CDF, then dropping to the horizontal axis, and picking where you landed to be your rv's realization. Think about doing this over and over. In the long run, think about the distribution of the realizations you will get.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 21:52
add a comment |
$begingroup$
Let $U$ be a random variable with uniform distribution on $(0,1)$.
Prescribe $Phi:(0,1)tomathbb R$ by:$$umapstoinf{xinmathbb Rmid uleq F(x)}$$
Then it can be shown that:$$Phi(u)leq xiff uleq F(x)$$
So letting $U$ be a random variable with uniform distribution on $(0,1)$ we have:$$Phi(U)leq xiff Uleq F(x)$$and consequently:$$F_{Phi(U)}(x)=P(Phi(u)leq x)=P(Uleq F(x))=F(x)$$
This always works.
In the special case where $F$ has an inverse we find that $Phi$ is actually that inverse.
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Let $U$ be a random variable with uniform distribution on $(0,1)$.
Prescribe $Phi:(0,1)tomathbb R$ by:$$umapstoinf{xinmathbb Rmid uleq F(x)}$$
Then it can be shown that:$$Phi(u)leq xiff uleq F(x)$$
So letting $U$ be a random variable with uniform distribution on $(0,1)$ we have:$$Phi(U)leq xiff Uleq F(x)$$and consequently:$$F_{Phi(U)}(x)=P(Phi(u)leq x)=P(Uleq F(x))=F(x)$$
This always works.
In the special case where $F$ has an inverse we find that $Phi$ is actually that inverse.
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Let $U$ be a random variable with uniform distribution on $(0,1)$.
Prescribe $Phi:(0,1)tomathbb R$ by:$$umapstoinf{xinmathbb Rmid uleq F(x)}$$
Then it can be shown that:$$Phi(u)leq xiff uleq F(x)$$
So letting $U$ be a random variable with uniform distribution on $(0,1)$ we have:$$Phi(U)leq xiff Uleq F(x)$$and consequently:$$F_{Phi(U)}(x)=P(Phi(u)leq x)=P(Uleq F(x))=F(x)$$
This always works.
In the special case where $F$ has an inverse we find that $Phi$ is actually that inverse.
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
$endgroup$
Let $U$ be a random variable with uniform distribution on $(0,1)$.
Prescribe $Phi:(0,1)tomathbb R$ by:$$umapstoinf{xinmathbb Rmid uleq F(x)}$$
Then it can be shown that:$$Phi(u)leq xiff uleq F(x)$$
So letting $U$ be a random variable with uniform distribution on $(0,1)$ we have:$$Phi(U)leq xiff Uleq F(x)$$and consequently:$$F_{Phi(U)}(x)=P(Phi(u)leq x)=P(Uleq F(x))=F(x)$$
This always works.
In the special case where $F$ has an inverse we find that $Phi$ is actually that inverse.
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
answered Aug 8 '17 at 13:05
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
If you are studying elementary probability theory, allow me to reformulate your question as "how can I represent a random variable $X$ with a given CDF $F_X$ in terms of a uniform random variable $U$ on $(0,1)$?" The answer to that is the quantile function: you define
$$G_X(p)=inf { x : F_X(x) geq p }$$
and then define $X$ to be $G_X(U)$.
Note that if $F_X$ is invertible then $G_X=F_X^{-1}$, otherwise this is "the right generalization". One can see this by looking at the discrete case: if $P(X=x)=p$ then $P(G_X(U)=x)=p$. This is because a jump of height $p$ in $F_X$ corresponds to a flat region of length $p$ in $G_X$, and the uniform distribution on $(0,1)$ assigns each interval a probability equal to its length.
The natural question is now "what's a uniform random variable on $(0,1)$?" Well, it has $F_U(x)=begin{cases} 0 & x<0 \ x & x in [0,1] \ 1 & x>1 end{cases}$. But otherwise such a thing is a black box from the elementary point of view.
If you are studying measure-theoretic probability theory then the answer is a bit more explicit. A random variable with CDF $F_X$ is given by $G_X : Omega to mathbb{R}$ where $G_X$ is the quantile function as defined before, $Omega=(0,1)$, $mathcal{F}$ is the Borel $sigma$-algebra on $(0,1)$, and $mathbb{P}$ is the Lebesgue measure. Note that on this space the identity function is a uniform random variable on $(0,1)$, so this is really the same construction as the one described above.
In any case these constructions can be generalized to finitely many random variables by looking at the uniform distribution on $(0,1)^n$ instead of $(0,1)$.
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
$endgroup$
1
$begingroup$
Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
add a comment |
$begingroup$
If you are studying elementary probability theory, allow me to reformulate your question as "how can I represent a random variable $X$ with a given CDF $F_X$ in terms of a uniform random variable $U$ on $(0,1)$?" The answer to that is the quantile function: you define
$$G_X(p)=inf { x : F_X(x) geq p }$$
and then define $X$ to be $G_X(U)$.
Note that if $F_X$ is invertible then $G_X=F_X^{-1}$, otherwise this is "the right generalization". One can see this by looking at the discrete case: if $P(X=x)=p$ then $P(G_X(U)=x)=p$. This is because a jump of height $p$ in $F_X$ corresponds to a flat region of length $p$ in $G_X$, and the uniform distribution on $(0,1)$ assigns each interval a probability equal to its length.
The natural question is now "what's a uniform random variable on $(0,1)$?" Well, it has $F_U(x)=begin{cases} 0 & x<0 \ x & x in [0,1] \ 1 & x>1 end{cases}$. But otherwise such a thing is a black box from the elementary point of view.
If you are studying measure-theoretic probability theory then the answer is a bit more explicit. A random variable with CDF $F_X$ is given by $G_X : Omega to mathbb{R}$ where $G_X$ is the quantile function as defined before, $Omega=(0,1)$, $mathcal{F}$ is the Borel $sigma$-algebra on $(0,1)$, and $mathbb{P}$ is the Lebesgue measure. Note that on this space the identity function is a uniform random variable on $(0,1)$, so this is really the same construction as the one described above.
In any case these constructions can be generalized to finitely many random variables by looking at the uniform distribution on $(0,1)^n$ instead of $(0,1)$.
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
$endgroup$
1
$begingroup$
Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
add a comment |
$begingroup$
If you are studying elementary probability theory, allow me to reformulate your question as "how can I represent a random variable $X$ with a given CDF $F_X$ in terms of a uniform random variable $U$ on $(0,1)$?" The answer to that is the quantile function: you define
$$G_X(p)=inf { x : F_X(x) geq p }$$
and then define $X$ to be $G_X(U)$.
Note that if $F_X$ is invertible then $G_X=F_X^{-1}$, otherwise this is "the right generalization". One can see this by looking at the discrete case: if $P(X=x)=p$ then $P(G_X(U)=x)=p$. This is because a jump of height $p$ in $F_X$ corresponds to a flat region of length $p$ in $G_X$, and the uniform distribution on $(0,1)$ assigns each interval a probability equal to its length.
The natural question is now "what's a uniform random variable on $(0,1)$?" Well, it has $F_U(x)=begin{cases} 0 & x<0 \ x & x in [0,1] \ 1 & x>1 end{cases}$. But otherwise such a thing is a black box from the elementary point of view.
If you are studying measure-theoretic probability theory then the answer is a bit more explicit. A random variable with CDF $F_X$ is given by $G_X : Omega to mathbb{R}$ where $G_X$ is the quantile function as defined before, $Omega=(0,1)$, $mathcal{F}$ is the Borel $sigma$-algebra on $(0,1)$, and $mathbb{P}$ is the Lebesgue measure. Note that on this space the identity function is a uniform random variable on $(0,1)$, so this is really the same construction as the one described above.
In any case these constructions can be generalized to finitely many random variables by looking at the uniform distribution on $(0,1)^n$ instead of $(0,1)$.
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
$endgroup$
If you are studying elementary probability theory, allow me to reformulate your question as "how can I represent a random variable $X$ with a given CDF $F_X$ in terms of a uniform random variable $U$ on $(0,1)$?" The answer to that is the quantile function: you define
$$G_X(p)=inf { x : F_X(x) geq p }$$
and then define $X$ to be $G_X(U)$.
Note that if $F_X$ is invertible then $G_X=F_X^{-1}$, otherwise this is "the right generalization". One can see this by looking at the discrete case: if $P(X=x)=p$ then $P(G_X(U)=x)=p$. This is because a jump of height $p$ in $F_X$ corresponds to a flat region of length $p$ in $G_X$, and the uniform distribution on $(0,1)$ assigns each interval a probability equal to its length.
The natural question is now "what's a uniform random variable on $(0,1)$?" Well, it has $F_U(x)=begin{cases} 0 & x<0 \ x & x in [0,1] \ 1 & x>1 end{cases}$. But otherwise such a thing is a black box from the elementary point of view.
If you are studying measure-theoretic probability theory then the answer is a bit more explicit. A random variable with CDF $F_X$ is given by $G_X : Omega to mathbb{R}$ where $G_X$ is the quantile function as defined before, $Omega=(0,1)$, $mathcal{F}$ is the Borel $sigma$-algebra on $(0,1)$, and $mathbb{P}$ is the Lebesgue measure. Note that on this space the identity function is a uniform random variable on $(0,1)$, so this is really the same construction as the one described above.
In any case these constructions can be generalized to finitely many random variables by looking at the uniform distribution on $(0,1)^n$ instead of $(0,1)$.
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
edited Sep 18 '17 at 15:55
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
answered Aug 8 '17 at 12:43
IanIan
68.8k25391
68.8k25391
1
$begingroup$
Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
add a comment |
1
$begingroup$
Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
1
1
$begingroup$
Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
$begingroup$
Would the downvoter care to comment? My guess would be that you are either criticizing the first paragraph or the longwindedness of the rest of it, but I am curious to hear your particular criticism, especially as it apparently does not pertain to any of the other answers.
$endgroup$
– Ian
Aug 8 '17 at 13:57
add a comment |
$begingroup$
The $text{cdf}$ of a random variable tells the probability that its value falls below a given bound,
$$text{cdf}_X(x)=mathbb P(X<x).$$
For a uniform variable $U$, we have
$$text{cdf}_U(x)=mathbb P(U<x)=x$$ for $xin[0,1]$.
Now if you want to create a random variable with an imposed $text{cdf}$, let $h(x)$, you want to achieve
$$mathbb P(X<x)=h(x).$$
If $X$ derives from a uniform variable via some transformation, let $X=g(U)$, you have
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x).$$
But also, assuming $g$ invertible, and using the fact that $U$ is uniform,
$$mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=g^{-1}(x).$$
Putting these facts together,
$$g^{-1}(x)=h(x)$$ or $$g(x)=h^{-1}(x).$$
For a simple example, take a negative exponential distribution,
$$text{cdf}_X=h(x)=1-e^{-x}.$$
Then by inversion,
$$g(x)=-ln(1-x)$$ which maps $0$ to $0$ and $1$ to $infty$. By looking at the plot of $g$, you will understand that the "density" of the transformed points goes decreasing, because the slope increases, resulting in a bias in favor of the small values.
We can even play the exercise when $U$ is not uniform but has a known $text{cdf}$, let $f(x)$. Then it suffices to adapt the equations, and
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x)=mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=f(g^{-1})(x)$$ and we need
$$g(x)=h^{-1}(f(x)).$$
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
$endgroup$
1
$begingroup$
The right-continuous convention for CDFs is much more common these days than the left-continuous one.
$endgroup$
– Ian
Aug 8 '17 at 13:19
$begingroup$
@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
1
$begingroup$
Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
add a comment |
$begingroup$
The $text{cdf}$ of a random variable tells the probability that its value falls below a given bound,
$$text{cdf}_X(x)=mathbb P(X<x).$$
For a uniform variable $U$, we have
$$text{cdf}_U(x)=mathbb P(U<x)=x$$ for $xin[0,1]$.
Now if you want to create a random variable with an imposed $text{cdf}$, let $h(x)$, you want to achieve
$$mathbb P(X<x)=h(x).$$
If $X$ derives from a uniform variable via some transformation, let $X=g(U)$, you have
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x).$$
But also, assuming $g$ invertible, and using the fact that $U$ is uniform,
$$mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=g^{-1}(x).$$
Putting these facts together,
$$g^{-1}(x)=h(x)$$ or $$g(x)=h^{-1}(x).$$
For a simple example, take a negative exponential distribution,
$$text{cdf}_X=h(x)=1-e^{-x}.$$
Then by inversion,
$$g(x)=-ln(1-x)$$ which maps $0$ to $0$ and $1$ to $infty$. By looking at the plot of $g$, you will understand that the "density" of the transformed points goes decreasing, because the slope increases, resulting in a bias in favor of the small values.
We can even play the exercise when $U$ is not uniform but has a known $text{cdf}$, let $f(x)$. Then it suffices to adapt the equations, and
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x)=mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=f(g^{-1})(x)$$ and we need
$$g(x)=h^{-1}(f(x)).$$
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
$endgroup$
1
$begingroup$
The right-continuous convention for CDFs is much more common these days than the left-continuous one.
$endgroup$
– Ian
Aug 8 '17 at 13:19
$begingroup$
@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
1
$begingroup$
Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
add a comment |
$begingroup$
The $text{cdf}$ of a random variable tells the probability that its value falls below a given bound,
$$text{cdf}_X(x)=mathbb P(X<x).$$
For a uniform variable $U$, we have
$$text{cdf}_U(x)=mathbb P(U<x)=x$$ for $xin[0,1]$.
Now if you want to create a random variable with an imposed $text{cdf}$, let $h(x)$, you want to achieve
$$mathbb P(X<x)=h(x).$$
If $X$ derives from a uniform variable via some transformation, let $X=g(U)$, you have
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x).$$
But also, assuming $g$ invertible, and using the fact that $U$ is uniform,
$$mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=g^{-1}(x).$$
Putting these facts together,
$$g^{-1}(x)=h(x)$$ or $$g(x)=h^{-1}(x).$$
For a simple example, take a negative exponential distribution,
$$text{cdf}_X=h(x)=1-e^{-x}.$$
Then by inversion,
$$g(x)=-ln(1-x)$$ which maps $0$ to $0$ and $1$ to $infty$. By looking at the plot of $g$, you will understand that the "density" of the transformed points goes decreasing, because the slope increases, resulting in a bias in favor of the small values.
We can even play the exercise when $U$ is not uniform but has a known $text{cdf}$, let $f(x)$. Then it suffices to adapt the equations, and
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x)=mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=f(g^{-1})(x)$$ and we need
$$g(x)=h^{-1}(f(x)).$$
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
$endgroup$
The $text{cdf}$ of a random variable tells the probability that its value falls below a given bound,
$$text{cdf}_X(x)=mathbb P(X<x).$$
For a uniform variable $U$, we have
$$text{cdf}_U(x)=mathbb P(U<x)=x$$ for $xin[0,1]$.
Now if you want to create a random variable with an imposed $text{cdf}$, let $h(x)$, you want to achieve
$$mathbb P(X<x)=h(x).$$
If $X$ derives from a uniform variable via some transformation, let $X=g(U)$, you have
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x).$$
But also, assuming $g$ invertible, and using the fact that $U$ is uniform,
$$mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=g^{-1}(x).$$
Putting these facts together,
$$g^{-1}(x)=h(x)$$ or $$g(x)=h^{-1}(x).$$
For a simple example, take a negative exponential distribution,
$$text{cdf}_X=h(x)=1-e^{-x}.$$
Then by inversion,
$$g(x)=-ln(1-x)$$ which maps $0$ to $0$ and $1$ to $infty$. By looking at the plot of $g$, you will understand that the "density" of the transformed points goes decreasing, because the slope increases, resulting in a bias in favor of the small values.
We can even play the exercise when $U$ is not uniform but has a known $text{cdf}$, let $f(x)$. Then it suffices to adapt the equations, and
$$mathbb P(X<x)=mathbb P(g(U)<x)=h(x)=mathbb P(g(U)<x)=mathbb P(U<g^{-1}(x))=f(g^{-1})(x)$$ and we need
$$g(x)=h^{-1}(f(x)).$$
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
edited Aug 8 '17 at 13:39
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
answered Aug 8 '17 at 13:17
Yves DaoustYves Daoust
130k676229
130k676229
1
$begingroup$
The right-continuous convention for CDFs is much more common these days than the left-continuous one.
$endgroup$
– Ian
Aug 8 '17 at 13:19
$begingroup$
@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
1
$begingroup$
Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
add a comment |
1
$begingroup$
The right-continuous convention for CDFs is much more common these days than the left-continuous one.
$endgroup$
– Ian
Aug 8 '17 at 13:19
$begingroup$
@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
1
$begingroup$
Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
1
1
$begingroup$
The right-continuous convention for CDFs is much more common these days than the left-continuous one.
$endgroup$
– Ian
Aug 8 '17 at 13:19
$begingroup$
The right-continuous convention for CDFs is much more common these days than the left-continuous one.
$endgroup$
– Ian
Aug 8 '17 at 13:19
$begingroup$
@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
$begingroup$
@Ian: yep but this is irrelevant to the present discussion.
$endgroup$
– Yves Daoust
Aug 8 '17 at 13:30
1
1
$begingroup$
Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
$begingroup$
Well...sort of. Of course everything turns out the same in either convention, but some definitions are altered in ways that might not be obvious. For example the quantile function is defined through a sup rather than an inf.
$endgroup$
– Ian
Aug 8 '17 at 13:32
add a comment |
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$begingroup$
I see that the answers discuss prob integral transform (en.wikipedia.org/wiki/Probability_integral_transform), which answers/clarifies the second paragraph of the question of generating a rv $X$ that has the given CDF from a standard uniform rv. But I don't know what the question in the title
given the cumulative distribution function find a random variable that has this distributionmeans since if a cdf $F(x)$ is given for some random variable $X$ then for any $a$ and $b$ the probability $P(a < X leq b)$ is nailed down by $F(b)-F(a)$ so $F(x)$ does indeed specify the rv a.e.$endgroup$
– Just_to_Answer
Aug 8 '17 at 16:16
$begingroup$
@Just_to_Answer Obviously F(x) is a random variable, but it's distribution isn't the same as F(x).
$endgroup$
– aroma
Aug 8 '17 at 17:31
$begingroup$
$F(x)$ by convention is not a random variable. It is a real valued function $mathbb R rightarrow [0,1]$ defined by $F(x) = P(X leq x)$. So my point is specifying $F(x)$ is a.e. same as specifying $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:41
$begingroup$
@Just_to_Answer I don't understand your point, even if $F(x)$ is a r.v. how does it help in finding a r.v. whose CDF is $F(x)$. Since CDF of the r.v. $F(x)$ is not necessarily $F(x)$.
$endgroup$
– aroma
Aug 8 '17 at 17:45
$begingroup$
Let me try it differently. You give me a valid CDF $F(x)$ and say this is the CDF of some rv $X$. Using just that $F(x)$ and nothing else, I can compute any probabilities, moments, etc.etc. involving the rv $X$.
$endgroup$
– Just_to_Answer
Aug 8 '17 at 17:52