Coin toss- expected value.
0
$begingroup$
A coin had tossed three times.
Let:
$X$-number of tails
$Y$-number of heads
Find the expected value and variance $Z=XY$
My solution:
$E(Z)=E(XY)= 2 cdot 1/4 + 3 cdot 1/4=6/4$
Because,
I know that $ X,Y=0,1,2,3$
so, $Z=2,3$, since $Z=0,1$ doesn't describe three throws.
And:
$E(Z^2)=13/4$
random-variables variance expected-value
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
$endgroup$
|
show 1 more comment
0
$begingroup$
A coin had tossed three times.
Let:
$X$-number of tails
$Y$-number of heads
Find the expected value and variance $Z=XY$
My solution:
$E(Z)=E(XY)= 2 cdot 1/4 + 3 cdot 1/4=6/4$
Because,
I know that $ X,Y=0,1,2,3$
so, $Z=2,3$, since $Z=0,1$ doesn't describe three throws.
And:
$E(Z^2)=13/4$
random-variables variance expected-value
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
$endgroup$
1
$begingroup$
Not following. How can $Z=3$?
$endgroup$
– lulu
Dec 11 '18 at 0:32
1
$begingroup$
You have $Y=3-X$, so $Z=X(3-X)$. This cannot take the value $3$ since $Xin{0,1,2,3}$. You cannot have $X=1$ and $Y=3$: since $X+Y=3$.
$endgroup$
– Clement C.
Dec 11 '18 at 0:35
$begingroup$
Right, so $Z=2$ and that's it.
$endgroup$
– PabloZ392
Dec 11 '18 at 0:39
$begingroup$
No. $Z$ can also be $0$. Also, if the $frac14's$ are meant to be probabilities, then I don't understand that either. They don't add to $1$, for example.
$endgroup$
– lulu
Dec 11 '18 at 0:40
1
$begingroup$
There's no reason for any confusion here. There are only $8$ cases and each has probability $frac 18$. Just list all of them and do the calculation by hand.
$endgroup$
– lulu
Dec 11 '18 at 0:41
|
show 1 more comment
0
0
0
$begingroup$
A coin had tossed three times.
Let:
$X$-number of tails
$Y$-number of heads
Find the expected value and variance $Z=XY$
My solution:
$E(Z)=E(XY)= 2 cdot 1/4 + 3 cdot 1/4=6/4$
Because,
I know that $ X,Y=0,1,2,3$
so, $Z=2,3$, since $Z=0,1$ doesn't describe three throws.
And:
$E(Z^2)=13/4$
random-variables variance expected-value
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
$endgroup$
A coin had tossed three times.
Let:
$X$-number of tails
$Y$-number of heads
Find the expected value and variance $Z=XY$
My solution:
$E(Z)=E(XY)= 2 cdot 1/4 + 3 cdot 1/4=6/4$
Because,
I know that $ X,Y=0,1,2,3$
so, $Z=2,3$, since $Z=0,1$ doesn't describe three throws.
And:
$E(Z^2)=13/4$
random-variables variance expected-value
random-variables variance expected-value
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
asked Dec 11 '18 at 0:29
PabloZ392PabloZ392
686
686
1
$begingroup$
Not following. How can $Z=3$?
$endgroup$
– lulu
Dec 11 '18 at 0:32
1
$begingroup$
You have $Y=3-X$, so $Z=X(3-X)$. This cannot take the value $3$ since $Xin{0,1,2,3}$. You cannot have $X=1$ and $Y=3$: since $X+Y=3$.
$endgroup$
– Clement C.
Dec 11 '18 at 0:35
$begingroup$
Right, so $Z=2$ and that's it.
$endgroup$
– PabloZ392
Dec 11 '18 at 0:39
$begingroup$
No. $Z$ can also be $0$. Also, if the $frac14's$ are meant to be probabilities, then I don't understand that either. They don't add to $1$, for example.
$endgroup$
– lulu
Dec 11 '18 at 0:40
1
$begingroup$
There's no reason for any confusion here. There are only $8$ cases and each has probability $frac 18$. Just list all of them and do the calculation by hand.
$endgroup$
– lulu
Dec 11 '18 at 0:41
|
show 1 more comment
1
$begingroup$
Not following. How can $Z=3$?
$endgroup$
– lulu
Dec 11 '18 at 0:32
1
$begingroup$
You have $Y=3-X$, so $Z=X(3-X)$. This cannot take the value $3$ since $Xin{0,1,2,3}$. You cannot have $X=1$ and $Y=3$: since $X+Y=3$.
$endgroup$
– Clement C.
Dec 11 '18 at 0:35
$begingroup$
Right, so $Z=2$ and that's it.
$endgroup$
– PabloZ392
Dec 11 '18 at 0:39
$begingroup$
No. $Z$ can also be $0$. Also, if the $frac14's$ are meant to be probabilities, then I don't understand that either. They don't add to $1$, for example.
$endgroup$
– lulu
Dec 11 '18 at 0:40
1
$begingroup$
There's no reason for any confusion here. There are only $8$ cases and each has probability $frac 18$. Just list all of them and do the calculation by hand.
$endgroup$
– lulu
Dec 11 '18 at 0:41
1
1
$begingroup$
Not following. How can $Z=3$?
$endgroup$
– lulu
Dec 11 '18 at 0:32
$begingroup$
Not following. How can $Z=3$?
$endgroup$
– lulu
Dec 11 '18 at 0:32
1
1
$begingroup$
You have $Y=3-X$, so $Z=X(3-X)$. This cannot take the value $3$ since $Xin{0,1,2,3}$. You cannot have $X=1$ and $Y=3$: since $X+Y=3$.
$endgroup$
– Clement C.
Dec 11 '18 at 0:35
$begingroup$
You have $Y=3-X$, so $Z=X(3-X)$. This cannot take the value $3$ since $Xin{0,1,2,3}$. You cannot have $X=1$ and $Y=3$: since $X+Y=3$.
$endgroup$
– Clement C.
Dec 11 '18 at 0:35
$begingroup$
Right, so $Z=2$ and that's it.
$endgroup$
– PabloZ392
Dec 11 '18 at 0:39
$begingroup$
Right, so $Z=2$ and that's it.
$endgroup$
– PabloZ392
Dec 11 '18 at 0:39
$begingroup$
No. $Z$ can also be $0$. Also, if the $frac14's$ are meant to be probabilities, then I don't understand that either. They don't add to $1$, for example.
$endgroup$
– lulu
Dec 11 '18 at 0:40
$begingroup$
No. $Z$ can also be $0$. Also, if the $frac14's$ are meant to be probabilities, then I don't understand that either. They don't add to $1$, for example.
$endgroup$
– lulu
Dec 11 '18 at 0:40
1
1
$begingroup$
There's no reason for any confusion here. There are only $8$ cases and each has probability $frac 18$. Just list all of them and do the calculation by hand.
$endgroup$
– lulu
Dec 11 '18 at 0:41
$begingroup$
There's no reason for any confusion here. There are only $8$ cases and each has probability $frac 18$. Just list all of them and do the calculation by hand.
$endgroup$
– lulu
Dec 11 '18 at 0:41
|
show 1 more 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
});
}
});
draft saved
draft discarded
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%2f3034699%2fcoin-toss-expected-value%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
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3034699%2fcoin-toss-expected-value%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
1
$begingroup$
Not following. How can $Z=3$?
$endgroup$
– lulu
Dec 11 '18 at 0:32
1
$begingroup$
You have $Y=3-X$, so $Z=X(3-X)$. This cannot take the value $3$ since $Xin{0,1,2,3}$. You cannot have $X=1$ and $Y=3$: since $X+Y=3$.
$endgroup$
– Clement C.
Dec 11 '18 at 0:35
$begingroup$
Right, so $Z=2$ and that's it.
$endgroup$
– PabloZ392
Dec 11 '18 at 0:39
$begingroup$
No. $Z$ can also be $0$. Also, if the $frac14's$ are meant to be probabilities, then I don't understand that either. They don't add to $1$, for example.
$endgroup$
– lulu
Dec 11 '18 at 0:40
1
$begingroup$
There's no reason for any confusion here. There are only $8$ cases and each has probability $frac 18$. Just list all of them and do the calculation by hand.
$endgroup$
– lulu
Dec 11 '18 at 0:41