Write the expression for $mathbb{E}(Y|X)$ The Next CEO of Stack OverflowMarkov Chain - Snakes...
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Write the expression for $mathbb{E}(Y|X)$
The Next CEO of Stack OverflowMarkov Chain - Snakes and LaddersExpected value of two successive heads or tails (stuck on computation)Conditional Probability of two random variablesExpected Value with stopping rule.Expectation of Coin Flipping GameConditional expectation for number of heads in a coin tossing depending on number shown on dieNumber of flips $X$ of a fair coin to win a gameConditional expectation in flipping coin. How much will I be willing to pay to see the first toss?Finding the probability of winning an interrupted game.Gambler's ruin problem variant with unfair coin and different starting funds?
$begingroup$
I have a confusion regarding how to go about solving the following question:
Suppose you are invited to play a game where your earnings are given by multiplying the outcome of rolling a fair die ($Z$) with tossing a fair coin ($X$). You have to pay $$5$ each time you want to play this game.
Let
$Y$ = net earnings,
$Z$ = outcome of dice ($1,2,ldots,6$),
$X$ = outcome of coin ($X$ equals $1$ if heads and $2$ if tails)
- Write $Y$ in terms of $X$ and $Z$
This comes out as $Y = XZ-5$ (I think)
Write the expression for $mathbb{E}(Y|X)$. What is the expected earnings if
we know the coin landed tailWhat is your expected earning from playing this game? Answer
without making a probability distribution for $Y$.
conditional-expectation conditional-probability
edited Mar 16 at 17:56
mwt
959416
asked Mar 16 at 9:49
FM109FM109
11
$endgroup$
add a comment |
$begingroup$
I have a confusion regarding how to go about solving the following question:
Suppose you are invited to play a game where your earnings are given by multiplying the outcome of rolling a fair die ($Z$) with tossing a fair coin ($X$). You have to pay $$5$ each time you want to play this game.
Let
$Y$ = net earnings,
$Z$ = outcome of dice ($1,2,ldots,6$),
$X$ = outcome of coin ($X$ equals $1$ if heads and $2$ if tails)
- Write $Y$ in terms of $X$ and $Z$
This comes out as $Y = XZ-5$ (I think)
Write the expression for $mathbb{E}(Y|X)$. What is the expected earnings if
we know the coin landed tailWhat is your expected earning from playing this game? Answer
without making a probability distribution for $Y$.
conditional-expectation conditional-probability
edited Mar 16 at 17:56
mwt
959416
asked Mar 16 at 9:49
FM109FM109
11
$endgroup$
$begingroup$
Kindly use Mathjax:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Mar 16 at 9:54
$begingroup$
Can you find an answer yourself for the question in the first bullet? Do an effort and edit.
$endgroup$
– drhab
Mar 16 at 10:41
$begingroup$
Numerical values of coin toss is a little arbitrary!. I can fix X to be 0 for tails and 1 for head and the net earnings ' equation is different. Has the problem not given what the numerical outcome of X?
$endgroup$
– Satish Ramanathan
Mar 16 at 12:00
$begingroup$
This is the entirety of the question I am afraid and it has me confused as to how I should go about it
$endgroup$
– FM109
Mar 16 at 12:49
add a comment |
$begingroup$
I have a confusion regarding how to go about solving the following question:
Suppose you are invited to play a game where your earnings are given by multiplying the outcome of rolling a fair die ($Z$) with tossing a fair coin ($X$). You have to pay $$5$ each time you want to play this game.
Let
$Y$ = net earnings,
$Z$ = outcome of dice ($1,2,ldots,6$),
$X$ = outcome of coin ($X$ equals $1$ if heads and $2$ if tails)
- Write $Y$ in terms of $X$ and $Z$
This comes out as $Y = XZ-5$ (I think)
Write the expression for $mathbb{E}(Y|X)$. What is the expected earnings if
we know the coin landed tailWhat is your expected earning from playing this game? Answer
without making a probability distribution for $Y$.
conditional-expectation conditional-probability
edited Mar 16 at 17:56
mwt
959416
asked Mar 16 at 9:49
FM109FM109
11
$endgroup$
I have a confusion regarding how to go about solving the following question:
Suppose you are invited to play a game where your earnings are given by multiplying the outcome of rolling a fair die ($Z$) with tossing a fair coin ($X$). You have to pay $$5$ each time you want to play this game.
Let
$Y$ = net earnings,
$Z$ = outcome of dice ($1,2,ldots,6$),
$X$ = outcome of coin ($X$ equals $1$ if heads and $2$ if tails)
- Write $Y$ in terms of $X$ and $Z$
This comes out as $Y = XZ-5$ (I think)
Write the expression for $mathbb{E}(Y|X)$. What is the expected earnings if
we know the coin landed tailWhat is your expected earning from playing this game? Answer
without making a probability distribution for $Y$.
conditional-expectation conditional-probability
conditional-expectation conditional-probability
edited Mar 16 at 17:56
mwt
959416
asked Mar 16 at 9:49
FM109FM109
11
edited Mar 16 at 17:56
mwt
959416
asked Mar 16 at 9:49
FM109FM109
11
edited Mar 16 at 17:56
mwt
959416
edited Mar 16 at 17:56
mwt
959416
edited Mar 16 at 17:56
mwt
959416
959416
asked Mar 16 at 9:49
FM109FM109
11
asked Mar 16 at 9:49
FM109FM109
11
asked Mar 16 at 9:49
FM109FM109
11
11
$begingroup$
Kindly use Mathjax:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Mar 16 at 9:54
$begingroup$
Can you find an answer yourself for the question in the first bullet? Do an effort and edit.
$endgroup$
– drhab
Mar 16 at 10:41
$begingroup$
Numerical values of coin toss is a little arbitrary!. I can fix X to be 0 for tails and 1 for head and the net earnings ' equation is different. Has the problem not given what the numerical outcome of X?
$endgroup$
– Satish Ramanathan
Mar 16 at 12:00
$begingroup$
This is the entirety of the question I am afraid and it has me confused as to how I should go about it
$endgroup$
– FM109
Mar 16 at 12:49
add a comment |
$begingroup$
Kindly use Mathjax:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Mar 16 at 9:54
$begingroup$
Can you find an answer yourself for the question in the first bullet? Do an effort and edit.
$endgroup$
– drhab
Mar 16 at 10:41
$begingroup$
Numerical values of coin toss is a little arbitrary!. I can fix X to be 0 for tails and 1 for head and the net earnings ' equation is different. Has the problem not given what the numerical outcome of X?
$endgroup$
– Satish Ramanathan
Mar 16 at 12:00
$begingroup$
This is the entirety of the question I am afraid and it has me confused as to how I should go about it
$endgroup$
– FM109
Mar 16 at 12:49
$begingroup$
Kindly use Mathjax:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Mar 16 at 9:54
$begingroup$
Kindly use Mathjax:math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Paras Khosla
Mar 16 at 9:54
$begingroup$
Can you find an answer yourself for the question in the first bullet? Do an effort and edit.
$endgroup$
– drhab
Mar 16 at 10:41
$begingroup$
Can you find an answer yourself for the question in the first bullet? Do an effort and edit.
$endgroup$
– drhab
Mar 16 at 10:41
$begingroup$
Numerical values of coin toss is a little arbitrary!. I can fix X to be 0 for tails and 1 for head and the net earnings ' equation is different. Has the problem not given what the numerical outcome of X?
$endgroup$
– Satish Ramanathan
Mar 16 at 12:00
$begingroup$
Numerical values of coin toss is a little arbitrary!. I can fix X to be 0 for tails and 1 for head and the net earnings ' equation is different. Has the problem not given what the numerical outcome of X?
$endgroup$
– Satish Ramanathan
Mar 16 at 12:00
$begingroup$
This is the entirety of the question I am afraid and it has me confused as to how I should go about it
$endgroup$
– FM109
Mar 16 at 12:49
$begingroup$
This is the entirety of the question I am afraid and it has me confused as to how I should go about it
$endgroup$
– FM109
Mar 16 at 12:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You just have to use the fact that dice roll and coin toss are independent, after that (and the fact that for independent variables $mathbb{E}(XY) = mathbb{E}(X) cdot mathbb{E}(Y)$) it's rather trivial.
edited Mar 16 at 17:56
mwt
959416
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
$endgroup$
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You just have to use the fact that dice roll and coin toss are independent, after that (and the fact that for independent variables $mathbb{E}(XY) = mathbb{E}(X) cdot mathbb{E}(Y)$) it's rather trivial.
edited Mar 16 at 17:56
mwt
959416
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
$endgroup$
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
add a comment |
$begingroup$
You just have to use the fact that dice roll and coin toss are independent, after that (and the fact that for independent variables $mathbb{E}(XY) = mathbb{E}(X) cdot mathbb{E}(Y)$) it's rather trivial.
edited Mar 16 at 17:56
mwt
959416
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
$endgroup$
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
add a comment |
$begingroup$
You just have to use the fact that dice roll and coin toss are independent, after that (and the fact that for independent variables $mathbb{E}(XY) = mathbb{E}(X) cdot mathbb{E}(Y)$) it's rather trivial.
edited Mar 16 at 17:56
mwt
959416
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
$endgroup$
You just have to use the fact that dice roll and coin toss are independent, after that (and the fact that for independent variables $mathbb{E}(XY) = mathbb{E}(X) cdot mathbb{E}(Y)$) it's rather trivial.
edited Mar 16 at 17:56
mwt
959416
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
edited Mar 16 at 17:56
mwt
959416
edited Mar 16 at 17:56
mwt
959416
edited Mar 16 at 17:56
mwt
959416
959416
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
answered Mar 16 at 10:07
Piotr PiękosPiotr Piękos
264
264
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
add a comment |
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
I am afraid I still do not understand your answer correctly
$endgroup$
– FM109
Mar 16 at 12:48
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
An answer will be helpful
$endgroup$
– FM109
Mar 17 at 3:51
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
well, you wrote Y in terms of X and Z correctly, now $mathbb{E}[Y] = mathbb{E}[XZ - 5] = mathbb{E}[XZ] - 5 = $(because X and Z are independent) $mathbb{E}[X] cdot mathbb{E}[Z] - 5.$ $mathbb{E}[Y|X]$ will be a function of expected value of Y given X. So you have to define a function g(x) that for a given coin toss result gives expected value of the whole experiment. So you "calculate E[Y] using x", where x is a constant, not a random variable. $g(x) = mathbb{E}[Y|X] = mathbb{E}[xZ - 5] = $(x is a number, it's important here that X and Z are independent)$ x cdot mathbb{E}[Z] - 5$.
$endgroup$
– Piotr Piękos
Mar 17 at 9:59
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
another simple way to go about $mathbb{E}[Y|X]$ without using independence is just writing explicitly writing the $mathbb{E}[Y|x=1]$ and $mathbb{E}[Y|x=2]$
$endgroup$
– Piotr Piękos
Mar 17 at 10:03
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
$begingroup$
This was really helpful, i had in mind that x will be a constant but your explanation has made it quite clear to me. Thanks a lot I really appreciate you taking out time to explain it neatly
$endgroup$
– FM109
Mar 20 at 4:56
add a comment |
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Kindly use Mathjax:math.meta.stackexchange.com/questions/5020/…
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– Paras Khosla
Mar 16 at 9:54
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Can you find an answer yourself for the question in the first bullet? Do an effort and edit.
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– drhab
Mar 16 at 10:41
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Numerical values of coin toss is a little arbitrary!. I can fix X to be 0 for tails and 1 for head and the net earnings ' equation is different. Has the problem not given what the numerical outcome of X?
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– Satish Ramanathan
Mar 16 at 12:00
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This is the entirety of the question I am afraid and it has me confused as to how I should go about it
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– FM109
Mar 16 at 12:49