Mixing
Calculating prediction using distribution mean/mode
$begingroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
60x^3(1-x)^2& 0 leq x leq 1\
0 & text{otherwise} \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
(a)
Pretty sure this is Beta Distribution $Xsimtext{Beta}(a = 4, b = 3)$. Which is $displaystyle E(X) = frac{a}{a+b}$. But, I am not sure how to do this considering the constant.
$$E(X) = 60int_{0}^{1} x^4(1-x)^2dx = 60 frac{Gamma(5)Gamma(3)}{Gamma(8)} = 60frac{4! 2 !}{7!} = frac{4}{7}$$
(b)
$f(x) = 60x^3(1-x)^2$
$f'(x) = 60 x^2 (3 - 8 x + 5 x^2) = 60x^2 (5x-3)(x-1)$
roots are $x = 0,1,3/5$. The mode is $x = 3/5$ since it gives the maximum value if you plug into original equation.
(c)
$text{MSE(mode)} = E(X - 3/5)^2 = E(X^2 - -6/5X + 9/25) = E(X^2) - 6/5E(X) + 9/25$
finding $displaystyle E(X^2) = 60int_{0}^{1} x^5(1-x)^2 dx = 60 frac{Gamma(6)Gamma(3)}{Gamma(9)} = 60 frac{5!2!}{8!} = 5/14$
so $E(X^2) - E(X) + 1/4 = 5/14 - 24/35 + 9/25 = 11/350$
Would this be corrected
EDIT: Fixed based on comments for b / c
probability probability-distributions
asked Jan 17 at 5:26
shahshah
404
$endgroup$
add a comment |
$begingroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
60x^3(1-x)^2& 0 leq x leq 1\
0 & text{otherwise} \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
(a)
Pretty sure this is Beta Distribution $Xsimtext{Beta}(a = 4, b = 3)$. Which is $displaystyle E(X) = frac{a}{a+b}$. But, I am not sure how to do this considering the constant.
$$E(X) = 60int_{0}^{1} x^4(1-x)^2dx = 60 frac{Gamma(5)Gamma(3)}{Gamma(8)} = 60frac{4! 2 !}{7!} = frac{4}{7}$$
(b)
$f(x) = 60x^3(1-x)^2$
$f'(x) = 60 x^2 (3 - 8 x + 5 x^2) = 60x^2 (5x-3)(x-1)$
roots are $x = 0,1,3/5$. The mode is $x = 3/5$ since it gives the maximum value if you plug into original equation.
(c)
$text{MSE(mode)} = E(X - 3/5)^2 = E(X^2 - -6/5X + 9/25) = E(X^2) - 6/5E(X) + 9/25$
finding $displaystyle E(X^2) = 60int_{0}^{1} x^5(1-x)^2 dx = 60 frac{Gamma(6)Gamma(3)}{Gamma(9)} = 60 frac{5!2!}{8!} = 5/14$
so $E(X^2) - E(X) + 1/4 = 5/14 - 24/35 + 9/25 = 11/350$
Would this be corrected
EDIT: Fixed based on comments for b / c
probability probability-distributions
asked Jan 17 at 5:26
shahshah
404
$endgroup$
$begingroup$
a) Integration should be fine. Not sure why you are not sure - if you are allowed to apply the formula directly, just apply it and plug in the numbers. b) Normally you should expand to $60(x^5 - 2x^4 + x^3)$ and then do differentiation. Actually I do not want to check the numerical calculations but when I see the computed mode is $1/2$, it must be wrong as $a neq b$, the beta distribution cannot be symmetric.If you apply the wiki formula you should get $3/5$ eventually. c) Methodology looks good. I think it is straight-forward enough that you do not need to seek any easier way to do.
$endgroup$
– BGM
Jan 17 at 7:21
add a comment |
$begingroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
60x^3(1-x)^2& 0 leq x leq 1\
0 & text{otherwise} \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
(a)
Pretty sure this is Beta Distribution $Xsimtext{Beta}(a = 4, b = 3)$. Which is $displaystyle E(X) = frac{a}{a+b}$. But, I am not sure how to do this considering the constant.
$$E(X) = 60int_{0}^{1} x^4(1-x)^2dx = 60 frac{Gamma(5)Gamma(3)}{Gamma(8)} = 60frac{4! 2 !}{7!} = frac{4}{7}$$
(b)
$f(x) = 60x^3(1-x)^2$
$f'(x) = 60 x^2 (3 - 8 x + 5 x^2) = 60x^2 (5x-3)(x-1)$
roots are $x = 0,1,3/5$. The mode is $x = 3/5$ since it gives the maximum value if you plug into original equation.
(c)
$text{MSE(mode)} = E(X - 3/5)^2 = E(X^2 - -6/5X + 9/25) = E(X^2) - 6/5E(X) + 9/25$
finding $displaystyle E(X^2) = 60int_{0}^{1} x^5(1-x)^2 dx = 60 frac{Gamma(6)Gamma(3)}{Gamma(9)} = 60 frac{5!2!}{8!} = 5/14$
so $E(X^2) - E(X) + 1/4 = 5/14 - 24/35 + 9/25 = 11/350$
Would this be corrected
EDIT: Fixed based on comments for b / c
probability probability-distributions
asked Jan 17 at 5:26
shahshah
404
$endgroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
60x^3(1-x)^2& 0 leq x leq 1\
0 & text{otherwise} \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
(a)
Pretty sure this is Beta Distribution $Xsimtext{Beta}(a = 4, b = 3)$. Which is $displaystyle E(X) = frac{a}{a+b}$. But, I am not sure how to do this considering the constant.
$$E(X) = 60int_{0}^{1} x^4(1-x)^2dx = 60 frac{Gamma(5)Gamma(3)}{Gamma(8)} = 60frac{4! 2 !}{7!} = frac{4}{7}$$
(b)
$f(x) = 60x^3(1-x)^2$
$f'(x) = 60 x^2 (3 - 8 x + 5 x^2) = 60x^2 (5x-3)(x-1)$
roots are $x = 0,1,3/5$. The mode is $x = 3/5$ since it gives the maximum value if you plug into original equation.
(c)
$text{MSE(mode)} = E(X - 3/5)^2 = E(X^2 - -6/5X + 9/25) = E(X^2) - 6/5E(X) + 9/25$
finding $displaystyle E(X^2) = 60int_{0}^{1} x^5(1-x)^2 dx = 60 frac{Gamma(6)Gamma(3)}{Gamma(9)} = 60 frac{5!2!}{8!} = 5/14$
so $E(X^2) - E(X) + 1/4 = 5/14 - 24/35 + 9/25 = 11/350$
Would this be corrected
EDIT: Fixed based on comments for b / c
probability probability-distributions
probability probability-distributions
asked Jan 17 at 5:26
shahshah
404
asked Jan 17 at 5:26
shahshah
404
edited Jan 18 at 4:22
shah
asked Jan 17 at 5:26
shahshah
404
asked Jan 17 at 5:26
shahshah
404
asked Jan 17 at 5:26
shahshah
404
404
$begingroup$
a) Integration should be fine. Not sure why you are not sure - if you are allowed to apply the formula directly, just apply it and plug in the numbers. b) Normally you should expand to $60(x^5 - 2x^4 + x^3)$ and then do differentiation. Actually I do not want to check the numerical calculations but when I see the computed mode is $1/2$, it must be wrong as $a neq b$, the beta distribution cannot be symmetric.If you apply the wiki formula you should get $3/5$ eventually. c) Methodology looks good. I think it is straight-forward enough that you do not need to seek any easier way to do.
$endgroup$
– BGM
Jan 17 at 7:21
add a comment |
$begingroup$
a) Integration should be fine. Not sure why you are not sure - if you are allowed to apply the formula directly, just apply it and plug in the numbers. b) Normally you should expand to $60(x^5 - 2x^4 + x^3)$ and then do differentiation. Actually I do not want to check the numerical calculations but when I see the computed mode is $1/2$, it must be wrong as $a neq b$, the beta distribution cannot be symmetric.If you apply the wiki formula you should get $3/5$ eventually. c) Methodology looks good. I think it is straight-forward enough that you do not need to seek any easier way to do.
$endgroup$
– BGM
Jan 17 at 7:21
$begingroup$
a) Integration should be fine. Not sure why you are not sure - if you are allowed to apply the formula directly, just apply it and plug in the numbers. b) Normally you should expand to $60(x^5 - 2x^4 + x^3)$ and then do differentiation. Actually I do not want to check the numerical calculations but when I see the computed mode is $1/2$, it must be wrong as $a neq b$, the beta distribution cannot be symmetric.If you apply the wiki formula you should get $3/5$ eventually. c) Methodology looks good. I think it is straight-forward enough that you do not need to seek any easier way to do.
$endgroup$
– BGM
Jan 17 at 7:21
$begingroup$
a) Integration should be fine. Not sure why you are not sure - if you are allowed to apply the formula directly, just apply it and plug in the numbers. b) Normally you should expand to $60(x^5 - 2x^4 + x^3)$ and then do differentiation. Actually I do not want to check the numerical calculations but when I see the computed mode is $1/2$, it must be wrong as $a neq b$, the beta distribution cannot be symmetric.If you apply the wiki formula you should get $3/5$ eventually. c) Methodology looks good. I think it is straight-forward enough that you do not need to seek any easier way to do.
$endgroup$
– BGM
Jan 17 at 7:21
add a comment |
0
active
oldest
votes
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%2f3076623%2fcalculating-prediction-using-distribution-mean-mode%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3076623%2fcalculating-prediction-using-distribution-mean-mode%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
$begingroup$
a) Integration should be fine. Not sure why you are not sure - if you are allowed to apply the formula directly, just apply it and plug in the numbers. b) Normally you should expand to $60(x^5 - 2x^4 + x^3)$ and then do differentiation. Actually I do not want to check the numerical calculations but when I see the computed mode is $1/2$, it must be wrong as $a neq b$, the beta distribution cannot be symmetric.If you apply the wiki formula you should get $3/5$ eventually. c) Methodology looks good. I think it is straight-forward enough that you do not need to seek any easier way to do.
$endgroup$
– BGM
Jan 17 at 7:21