Parameters of normal distribution: why are they so different from other distributions?
$begingroup$
In all, at least well-known probabilities, the parameters are not related to mean and variance. In other words, usually we provide a specification on how a single events occur (e.g., p in Bernoulli trial, frequency in Poisson / exponential). Then, we can calculate the expected value and variance. In contrast, in Normal distribution we provide mean (which is expected value) and variance (standard deviation). So, this looks as though this is not an information about single events.
Why is this so different for Normal distribution?
Thanks
probability probability-distributions normal-distribution
asked Dec 23 '18 at 17:05
JohnJohn
1236
$endgroup$
add a comment |
$begingroup$
In all, at least well-known probabilities, the parameters are not related to mean and variance. In other words, usually we provide a specification on how a single events occur (e.g., p in Bernoulli trial, frequency in Poisson / exponential). Then, we can calculate the expected value and variance. In contrast, in Normal distribution we provide mean (which is expected value) and variance (standard deviation). So, this looks as though this is not an information about single events.
Why is this so different for Normal distribution?
Thanks
probability probability-distributions normal-distribution
asked Dec 23 '18 at 17:05
JohnJohn
1236
$endgroup$
$begingroup$
It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter.
$endgroup$
– snar
Dec 23 '18 at 18:18
$begingroup$
Thanks for the answer. This sounds reasonable.
$endgroup$
– John
Dec 24 '18 at 6:37
add a comment |
$begingroup$
In all, at least well-known probabilities, the parameters are not related to mean and variance. In other words, usually we provide a specification on how a single events occur (e.g., p in Bernoulli trial, frequency in Poisson / exponential). Then, we can calculate the expected value and variance. In contrast, in Normal distribution we provide mean (which is expected value) and variance (standard deviation). So, this looks as though this is not an information about single events.
Why is this so different for Normal distribution?
Thanks
probability probability-distributions normal-distribution
asked Dec 23 '18 at 17:05
JohnJohn
1236
$endgroup$
In all, at least well-known probabilities, the parameters are not related to mean and variance. In other words, usually we provide a specification on how a single events occur (e.g., p in Bernoulli trial, frequency in Poisson / exponential). Then, we can calculate the expected value and variance. In contrast, in Normal distribution we provide mean (which is expected value) and variance (standard deviation). So, this looks as though this is not an information about single events.
Why is this so different for Normal distribution?
Thanks
probability probability-distributions normal-distribution
probability probability-distributions normal-distribution
asked Dec 23 '18 at 17:05
JohnJohn
1236
asked Dec 23 '18 at 17:05
JohnJohn
1236
asked Dec 23 '18 at 17:05
JohnJohn
1236
asked Dec 23 '18 at 17:05
JohnJohn
1236
asked Dec 23 '18 at 17:05
JohnJohn
1236
1236
$begingroup$
It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter.
$endgroup$
– snar
Dec 23 '18 at 18:18
$begingroup$
Thanks for the answer. This sounds reasonable.
$endgroup$
– John
Dec 24 '18 at 6:37
add a comment |
$begingroup$
It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter.
$endgroup$
– snar
Dec 23 '18 at 18:18
$begingroup$
Thanks for the answer. This sounds reasonable.
$endgroup$
– John
Dec 24 '18 at 6:37
$begingroup$
It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter.
$endgroup$
– snar
Dec 23 '18 at 18:18
$begingroup$
It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter.
$endgroup$
– snar
Dec 23 '18 at 18:18
$begingroup$
Thanks for the answer. This sounds reasonable.
$endgroup$
– John
Dec 24 '18 at 6:37
$begingroup$
Thanks for the answer. This sounds reasonable.
$endgroup$
– John
Dec 24 '18 at 6:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Normal distribution is a continuous distribution, while the others you mentioned (Bernoulli, Poisson) are discrete.
For continuous distributions, the probability of a single event is $0$. The discussion will always be about events within a range.
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
$endgroup$
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
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%2f3050500%2fparameters-of-normal-distribution-why-are-they-so-different-from-other-distribu%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
$begingroup$
Normal distribution is a continuous distribution, while the others you mentioned (Bernoulli, Poisson) are discrete.
For continuous distributions, the probability of a single event is $0$. The discussion will always be about events within a range.
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
$endgroup$
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
add a comment |
$begingroup$
Normal distribution is a continuous distribution, while the others you mentioned (Bernoulli, Poisson) are discrete.
For continuous distributions, the probability of a single event is $0$. The discussion will always be about events within a range.
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
$endgroup$
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
add a comment |
$begingroup$
Normal distribution is a continuous distribution, while the others you mentioned (Bernoulli, Poisson) are discrete.
For continuous distributions, the probability of a single event is $0$. The discussion will always be about events within a range.
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
$endgroup$
Normal distribution is a continuous distribution, while the others you mentioned (Bernoulli, Poisson) are discrete.
For continuous distributions, the probability of a single event is $0$. The discussion will always be about events within a range.
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
answered Dec 23 '18 at 17:10
herb steinbergherb steinberg
2,8732310
2,8732310
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
add a comment |
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution.
$endgroup$
– John
Dec 23 '18 at 17:27
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean).
$endgroup$
– herb steinberg
Dec 23 '18 at 17:43
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
But put it other way, why there are no distributions with variance as a parameter?
$endgroup$
– John
Dec 23 '18 at 18:04
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
$begingroup$
Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$.
$endgroup$
– snar
Dec 23 '18 at 18:22
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%2f3050500%2fparameters-of-normal-distribution-why-are-they-so-different-from-other-distribu%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
$begingroup$
It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter.
$endgroup$
– snar
Dec 23 '18 at 18:18
$begingroup$
Thanks for the answer. This sounds reasonable.
$endgroup$
– John
Dec 24 '18 at 6:37