Is $P(U=u)=1/x$ well defined when $x$ is the values a discrete random variable takes
$begingroup$
If $X$ is a discrete random variable with the following probability distribution function
$P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.
Now let $U$ be a discrete uniform random variable such that :
$P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)
I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
How can the probability that a random variable $U$ take the value $u$ be random?
If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?
Thanks
probability-theory probability-distributions
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
$endgroup$
add a comment |
$begingroup$
If $X$ is a discrete random variable with the following probability distribution function
$P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.
Now let $U$ be a discrete uniform random variable such that :
$P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)
I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
How can the probability that a random variable $U$ take the value $u$ be random?
If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?
Thanks
probability-theory probability-distributions
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
$endgroup$
add a comment |
$begingroup$
If $X$ is a discrete random variable with the following probability distribution function
$P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.
Now let $U$ be a discrete uniform random variable such that :
$P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)
I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
How can the probability that a random variable $U$ take the value $u$ be random?
If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?
Thanks
probability-theory probability-distributions
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
$endgroup$
If $X$ is a discrete random variable with the following probability distribution function
$P(X=x)=c(x-1),x=2,3,4,5,6$ where $c$ is the normalizing constant.
Now let $U$ be a discrete uniform random variable such that :
$P(U=u)=1/x,u=1,2,dots ,x( x$ is the value taken by $X$)
I was wondering if $P(U=u)$ is well-defined in the sense that $P(U=u)=1/x$ defined above makes sense?
How can the probability that a random variable $U$ take the value $u$ be random?
If I fix $x=2$ then $P(U=1)=P(U=2)=1/2$ , but then for $X=3$ we get $P(U=1)=P(U=2)=P(U=3)=1/3$ . How can we reconcile the two?
Thanks
probability-theory probability-distributions
probability-theory probability-distributions
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
asked Jan 7 at 20:49
user3503589user3503589
1,2351821
1,2351821
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$
Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.
To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$
answered Jan 7 at 21:18
lulululu
39.8k24778
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065473%2fis-pu-u-1-x-well-defined-when-x-is-the-values-a-discrete-random-variable-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$
Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.
To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$
answered Jan 7 at 21:18
lulululu
39.8k24778
$endgroup$
add a comment |
$begingroup$
I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$
Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.
To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$
answered Jan 7 at 21:18
lulululu
39.8k24778
$endgroup$
add a comment |
$begingroup$
I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$
Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.
To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$
answered Jan 7 at 21:18
lulululu
39.8k24778
$endgroup$
I'd read this as a conditional probability, namely: $$P(U=u,|,X=x)=frac 1x$$
Note that, given $X=x$, $U$ is permitted to take $x$ values so the sum of the conditional probabilities is $1$ as it should be.
To gain intuition for the set up, imagine that you first throw a fair die and record its value $X$. Then you toss a coin $X$ times and let $Y$ be the number of Heads you see. Then of course the distribution of $Y$ depends on $X$. For example:$$P(Y=0,|,X=3)=frac 1{2^3}$$ while $$P(Y=0,|,X=6)=frac 1{2^6}$$
and, in general, $$P(Y=0,|,X=x)=frac 1{2^x}$$
answered Jan 7 at 21:18
lulululu
39.8k24778
edited Jan 7 at 21:37
answered Jan 7 at 21:18
lulululu
39.8k24778
answered Jan 7 at 21:18
lulululu
39.8k24778
answered Jan 7 at 21:18
lulululu
39.8k24778
39.8k24778
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065473%2fis-pu-u-1-x-well-defined-when-x-is-the-values-a-discrete-random-variable-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
