Mathematical expectation question
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Following are the results obtained for two independent random variable x and y. $E(x)=4,E(y)=6,v(x)=5$ and $V(y)=4$, so find the value of
$1) $ $E(2x-y)^2$
$2) $ $E(7-2x-5y)$
That's the way I am trying :
$$V(x)=E(x^2)-E(x)^2 $$
$$5=E(x^2)-16$$
$$E(x^2)=21$$
In the same way
$$E(y^2)=40$$
$$E(xy)=6times 4=24$$
beginalignE(2x-y)^2 &=4E(x^2)-4E(xy)+E(y^2) \
&=4(21)-4(24)+40\
&=84-96+40\
&=28
endalign
But actual answer is $60$ which doesn't match.
discrete-mathematics
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up vote
0
down vote
favorite
Following are the results obtained for two independent random variable x and y. $E(x)=4,E(y)=6,v(x)=5$ and $V(y)=4$, so find the value of
$1) $ $E(2x-y)^2$
$2) $ $E(7-2x-5y)$
That's the way I am trying :
$$V(x)=E(x^2)-E(x)^2 $$
$$5=E(x^2)-16$$
$$E(x^2)=21$$
In the same way
$$E(y^2)=40$$
$$E(xy)=6times 4=24$$
beginalignE(2x-y)^2 &=4E(x^2)-4E(xy)+E(y^2) \
&=4(21)-4(24)+40\
&=84-96+40\
&=28
endalign
But actual answer is $60$ which doesn't match.
discrete-mathematics
As you may have noticed, most of your posts was closed down. Do not just paste the question statement as shown in a typical homework format on the site. Include your thoughts and attempts. It is good to include your working in the original post as compared to telling us what you have found without showing us the working. Take it as an exercise to edit the post to include the working so that one of us can tell where you make the mistake if there is any. Also, textbooks do make mistakes.
â Siong Thye Goh
Aug 27 at 15:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Following are the results obtained for two independent random variable x and y. $E(x)=4,E(y)=6,v(x)=5$ and $V(y)=4$, so find the value of
$1) $ $E(2x-y)^2$
$2) $ $E(7-2x-5y)$
That's the way I am trying :
$$V(x)=E(x^2)-E(x)^2 $$
$$5=E(x^2)-16$$
$$E(x^2)=21$$
In the same way
$$E(y^2)=40$$
$$E(xy)=6times 4=24$$
beginalignE(2x-y)^2 &=4E(x^2)-4E(xy)+E(y^2) \
&=4(21)-4(24)+40\
&=84-96+40\
&=28
endalign
But actual answer is $60$ which doesn't match.
discrete-mathematics
Following are the results obtained for two independent random variable x and y. $E(x)=4,E(y)=6,v(x)=5$ and $V(y)=4$, so find the value of
$1) $ $E(2x-y)^2$
$2) $ $E(7-2x-5y)$
That's the way I am trying :
$$V(x)=E(x^2)-E(x)^2 $$
$$5=E(x^2)-16$$
$$E(x^2)=21$$
In the same way
$$E(y^2)=40$$
$$E(xy)=6times 4=24$$
beginalignE(2x-y)^2 &=4E(x^2)-4E(xy)+E(y^2) \
&=4(21)-4(24)+40\
&=84-96+40\
&=28
endalign
But actual answer is $60$ which doesn't match.
discrete-mathematics
edited Aug 27 at 16:35
Siong Thye Goh
81k1453102
81k1453102
asked Aug 27 at 15:06
Aayush Mukharji
214
214
As you may have noticed, most of your posts was closed down. Do not just paste the question statement as shown in a typical homework format on the site. Include your thoughts and attempts. It is good to include your working in the original post as compared to telling us what you have found without showing us the working. Take it as an exercise to edit the post to include the working so that one of us can tell where you make the mistake if there is any. Also, textbooks do make mistakes.
â Siong Thye Goh
Aug 27 at 15:43
add a comment |Â
As you may have noticed, most of your posts was closed down. Do not just paste the question statement as shown in a typical homework format on the site. Include your thoughts and attempts. It is good to include your working in the original post as compared to telling us what you have found without showing us the working. Take it as an exercise to edit the post to include the working so that one of us can tell where you make the mistake if there is any. Also, textbooks do make mistakes.
â Siong Thye Goh
Aug 27 at 15:43
As you may have noticed, most of your posts was closed down. Do not just paste the question statement as shown in a typical homework format on the site. Include your thoughts and attempts. It is good to include your working in the original post as compared to telling us what you have found without showing us the working. Take it as an exercise to edit the post to include the working so that one of us can tell where you make the mistake if there is any. Also, textbooks do make mistakes.
â Siong Thye Goh
Aug 27 at 15:43
As you may have noticed, most of your posts was closed down. Do not just paste the question statement as shown in a typical homework format on the site. Include your thoughts and attempts. It is good to include your working in the original post as compared to telling us what you have found without showing us the working. Take it as an exercise to edit the post to include the working so that one of us can tell where you make the mistake if there is any. Also, textbooks do make mistakes.
â Siong Thye Goh
Aug 27 at 15:43
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
For the first part,beginalign
E[(2X-Y)^2] &= Var((2X-Y)) + [E(2X-Y)]^2 \
&= 4Var(X)+Var(Y)+(2E(X)-E(Y))^2
endalign
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
add a comment |Â
up vote
0
down vote
Hint: For independent random variables, $E[XY]=E[X]cdot E[Y]$.
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
For the first part,beginalign
E[(2X-Y)^2] &= Var((2X-Y)) + [E(2X-Y)]^2 \
&= 4Var(X)+Var(Y)+(2E(X)-E(Y))^2
endalign
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
add a comment |Â
up vote
0
down vote
accepted
For the first part,beginalign
E[(2X-Y)^2] &= Var((2X-Y)) + [E(2X-Y)]^2 \
&= 4Var(X)+Var(Y)+(2E(X)-E(Y))^2
endalign
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
For the first part,beginalign
E[(2X-Y)^2] &= Var((2X-Y)) + [E(2X-Y)]^2 \
&= 4Var(X)+Var(Y)+(2E(X)-E(Y))^2
endalign
For the first part,beginalign
E[(2X-Y)^2] &= Var((2X-Y)) + [E(2X-Y)]^2 \
&= 4Var(X)+Var(Y)+(2E(X)-E(Y))^2
endalign
answered Aug 27 at 15:40
Siong Thye Goh
81k1453102
81k1453102
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
add a comment |Â
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
I did this way answer comes 28. Is it correct or not please tell me.Because my answer is not matching to the book answer .
â Aayush Mukharji
Aug 27 at 16:28
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
your answer is correct.
â Siong Thye Goh
Aug 27 at 16:32
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
Thanks. But lot of people writing in comments that my questions are not asked in prescribed format and I am not showing my workings. Actually I am new to this site. Can you guide me about asking questions.
â Aayush Mukharji
Aug 27 at 16:34
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
what most people are looking for is what you have attempted so far. Here is a link on how to ask a question on the site. Most of the time just try your best to include your working in mathjax. It is alright if your attempt is wrong. Welcome to MSE.
â Siong Thye Goh
Aug 27 at 16:38
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
Thanks you very much you are really kind and helpful.
â Aayush Mukharji
Aug 27 at 16:40
add a comment |Â
up vote
0
down vote
Hint: For independent random variables, $E[XY]=E[X]cdot E[Y]$.
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
add a comment |Â
up vote
0
down vote
Hint: For independent random variables, $E[XY]=E[X]cdot E[Y]$.
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: For independent random variables, $E[XY]=E[X]cdot E[Y]$.
Hint: For independent random variables, $E[XY]=E[X]cdot E[Y]$.
answered Aug 27 at 15:11
paw88789
28.3k12248
28.3k12248
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
add a comment |Â
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
I tried this but my answer comes 28 which doesn't match the actual answer is it me wrong or actual answer?
â Aayush Mukharji
Aug 27 at 15:16
add a comment |Â
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As you may have noticed, most of your posts was closed down. Do not just paste the question statement as shown in a typical homework format on the site. Include your thoughts and attempts. It is good to include your working in the original post as compared to telling us what you have found without showing us the working. Take it as an exercise to edit the post to include the working so that one of us can tell where you make the mistake if there is any. Also, textbooks do make mistakes.
â Siong Thye Goh
Aug 27 at 15:43