Probability for Sampling Schemes
I'm currently doing my homework for my stats class and I have a question for one of my exercises:
Consider a population P of size N. We define a new sampling scheme in the following way: we first select a sample SA using a simple random sampling without replacement of size n1. Then, we sample SB in P outside of SA according to a simple random sampling without replacement of size n2. We obtain in a such a way the final sample S as S = SA ∪ SB.
Obtain the probability P(S = s) for a realization s (probability mass function). Is it equivalent to a simple random sampling with sample size n = n1 + n2?
My questions is:
I don't really understand the question and don't know what to answer. What kind of PDF am I supposed to describe? What does he mean by P(S=s) for some realization s?
In my understanding the question is asking what the probability is that the sample will contain a certain random variable. Can someone maybe give me a more concrete example so I can work it out?
probability statistics
add a comment |
I'm currently doing my homework for my stats class and I have a question for one of my exercises:
Consider a population P of size N. We define a new sampling scheme in the following way: we first select a sample SA using a simple random sampling without replacement of size n1. Then, we sample SB in P outside of SA according to a simple random sampling without replacement of size n2. We obtain in a such a way the final sample S as S = SA ∪ SB.
Obtain the probability P(S = s) for a realization s (probability mass function). Is it equivalent to a simple random sampling with sample size n = n1 + n2?
My questions is:
I don't really understand the question and don't know what to answer. What kind of PDF am I supposed to describe? What does he mean by P(S=s) for some realization s?
In my understanding the question is asking what the probability is that the sample will contain a certain random variable. Can someone maybe give me a more concrete example so I can work it out?
probability statistics
add a comment |
I'm currently doing my homework for my stats class and I have a question for one of my exercises:
Consider a population P of size N. We define a new sampling scheme in the following way: we first select a sample SA using a simple random sampling without replacement of size n1. Then, we sample SB in P outside of SA according to a simple random sampling without replacement of size n2. We obtain in a such a way the final sample S as S = SA ∪ SB.
Obtain the probability P(S = s) for a realization s (probability mass function). Is it equivalent to a simple random sampling with sample size n = n1 + n2?
My questions is:
I don't really understand the question and don't know what to answer. What kind of PDF am I supposed to describe? What does he mean by P(S=s) for some realization s?
In my understanding the question is asking what the probability is that the sample will contain a certain random variable. Can someone maybe give me a more concrete example so I can work it out?
probability statistics
I'm currently doing my homework for my stats class and I have a question for one of my exercises:
Consider a population P of size N. We define a new sampling scheme in the following way: we first select a sample SA using a simple random sampling without replacement of size n1. Then, we sample SB in P outside of SA according to a simple random sampling without replacement of size n2. We obtain in a such a way the final sample S as S = SA ∪ SB.
Obtain the probability P(S = s) for a realization s (probability mass function). Is it equivalent to a simple random sampling with sample size n = n1 + n2?
My questions is:
I don't really understand the question and don't know what to answer. What kind of PDF am I supposed to describe? What does he mean by P(S=s) for some realization s?
In my understanding the question is asking what the probability is that the sample will contain a certain random variable. Can someone maybe give me a more concrete example so I can work it out?
probability statistics
probability statistics
edited Nov 24 at 23:35
amWhy
191k28224439
191k28224439
asked Nov 24 at 22:52
Nicola Zaugg
41
41
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Unless I am interpreting wrong, here is my translation of the problem
In a population of size $P$, take a simple random sample of size $n_1$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $n_2$. Let our sample $S$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $n_1 + n_2$ from the original population of size $P$ ?
A realization is, put simply, an observed value
For $mathbb{P}(S=s)$, remember what $S$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $s$ (little s) could be any particular set of objects from the population.
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
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%2f3012199%2fprobability-for-sampling-schemes%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
Unless I am interpreting wrong, here is my translation of the problem
In a population of size $P$, take a simple random sample of size $n_1$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $n_2$. Let our sample $S$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $n_1 + n_2$ from the original population of size $P$ ?
A realization is, put simply, an observed value
For $mathbb{P}(S=s)$, remember what $S$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $s$ (little s) could be any particular set of objects from the population.
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
add a comment |
Unless I am interpreting wrong, here is my translation of the problem
In a population of size $P$, take a simple random sample of size $n_1$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $n_2$. Let our sample $S$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $n_1 + n_2$ from the original population of size $P$ ?
A realization is, put simply, an observed value
For $mathbb{P}(S=s)$, remember what $S$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $s$ (little s) could be any particular set of objects from the population.
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
add a comment |
Unless I am interpreting wrong, here is my translation of the problem
In a population of size $P$, take a simple random sample of size $n_1$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $n_2$. Let our sample $S$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $n_1 + n_2$ from the original population of size $P$ ?
A realization is, put simply, an observed value
For $mathbb{P}(S=s)$, remember what $S$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $s$ (little s) could be any particular set of objects from the population.
Unless I am interpreting wrong, here is my translation of the problem
In a population of size $P$, take a simple random sample of size $n_1$ without replacement. Next, without putting anything from this first sample back, take a second simple random sample of size $n_2$. Let our sample $S$ be the union of these two smaller samples. Is this the same as taking one large simple random sample of size $n_1 + n_2$ from the original population of size $P$ ?
A realization is, put simply, an observed value
For $mathbb{P}(S=s)$, remember what $S$ (big s) represents. It is defined as the union of two disjoint samples, so it would be a set of objects. $s$ (little s) could be any particular set of objects from the population.
answered Nov 24 at 23:37
WaveX
2,4902721
2,4902721
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
add a comment |
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
thanks for your answer. I see how the sampling scheme works, but how do I put this formally into a probability mass function since i dont know what s is exactly. My attempt so far was to show that either way you can make the same amount of distinct samples, which implies that the probability is the same. But how can i exactly show that with a probability mass function?
– Nicola Zaugg
Nov 25 at 12:04
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
We have $n_1+n_2$ objects being chosen from $P$ total objects as our sample. So there are ${P}choose{n_1+n_2}$ different possible samples we could have. And if each sample is equally likely, the probability for any particular sample being selected would be $1 / {P}choose{n_1+n_2}$
– WaveX
Nov 25 at 16:47
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3012199%2fprobability-for-sampling-schemes%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