Expected Value: How to determine using random variables that these E(X) equations are equivalent or not?
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To determine for each scenario that if for any random variables X, Y, Z that these are true or false: All of them are false but how do I prove it? 3) looks like it should be true if I consider a coin toss scenario, same with 2). How are all these false??
1) $E(min(X,Y,Z)) = min(E(X), E(Y), E(Z)) $
2) $E(X.Y) = E(X).E(Y)$
3) $E(1/X) = 1/E(X)$
discrete-mathematics random-variables expected-value
asked Dec 18 '18 at 4:45
TobyToby
1577
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add a comment |
$begingroup$
To determine for each scenario that if for any random variables X, Y, Z that these are true or false: All of them are false but how do I prove it? 3) looks like it should be true if I consider a coin toss scenario, same with 2). How are all these false??
1) $E(min(X,Y,Z)) = min(E(X), E(Y), E(Z)) $
2) $E(X.Y) = E(X).E(Y)$
3) $E(1/X) = 1/E(X)$
discrete-mathematics random-variables expected-value
asked Dec 18 '18 at 4:45
TobyToby
1577
$endgroup$
$begingroup$
For 3: Just take an example coin toss scenario and compute those values. There are only two choices for $X$ (0 or 1). Or make it 1 or 2 if you prefer. All you need is one particular example to show the equality does not hold. It would be unfortunate if you end up not trying any examples at all.
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– Michael
Dec 18 '18 at 5:12
add a comment |
$begingroup$
To determine for each scenario that if for any random variables X, Y, Z that these are true or false: All of them are false but how do I prove it? 3) looks like it should be true if I consider a coin toss scenario, same with 2). How are all these false??
1) $E(min(X,Y,Z)) = min(E(X), E(Y), E(Z)) $
2) $E(X.Y) = E(X).E(Y)$
3) $E(1/X) = 1/E(X)$
discrete-mathematics random-variables expected-value
asked Dec 18 '18 at 4:45
TobyToby
1577
$endgroup$
To determine for each scenario that if for any random variables X, Y, Z that these are true or false: All of them are false but how do I prove it? 3) looks like it should be true if I consider a coin toss scenario, same with 2). How are all these false??
1) $E(min(X,Y,Z)) = min(E(X), E(Y), E(Z)) $
2) $E(X.Y) = E(X).E(Y)$
3) $E(1/X) = 1/E(X)$
discrete-mathematics random-variables expected-value
discrete-mathematics random-variables expected-value
asked Dec 18 '18 at 4:45
TobyToby
1577
asked Dec 18 '18 at 4:45
TobyToby
1577
asked Dec 18 '18 at 4:45
TobyToby
1577
asked Dec 18 '18 at 4:45
TobyToby
1577
asked Dec 18 '18 at 4:45
TobyToby
1577
1577
$begingroup$
For 3: Just take an example coin toss scenario and compute those values. There are only two choices for $X$ (0 or 1). Or make it 1 or 2 if you prefer. All you need is one particular example to show the equality does not hold. It would be unfortunate if you end up not trying any examples at all.
$endgroup$
– Michael
Dec 18 '18 at 5:12
add a comment |
$begingroup$
For 3: Just take an example coin toss scenario and compute those values. There are only two choices for $X$ (0 or 1). Or make it 1 or 2 if you prefer. All you need is one particular example to show the equality does not hold. It would be unfortunate if you end up not trying any examples at all.
$endgroup$
– Michael
Dec 18 '18 at 5:12
$begingroup$
For 3: Just take an example coin toss scenario and compute those values. There are only two choices for $X$ (0 or 1). Or make it 1 or 2 if you prefer. All you need is one particular example to show the equality does not hold. It would be unfortunate if you end up not trying any examples at all.
$endgroup$
– Michael
Dec 18 '18 at 5:12
$begingroup$
For 3: Just take an example coin toss scenario and compute those values. There are only two choices for $X$ (0 or 1). Or make it 1 or 2 if you prefer. All you need is one particular example to show the equality does not hold. It would be unfortunate if you end up not trying any examples at all.
$endgroup$
– Michael
Dec 18 '18 at 5:12
add a comment |
1 Answer
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To see that they are false, all you need to do is provide a counterexample for each, ideally one for which the LHS and RHS can be easily computed.
Here are some suggestions:
Take $X, Y$ degenerate and both equal to zero with probability one. Can you think of a $Z$ which has expected value zero (so the RHS is zero) but for which the LHS is negative? (You can choose a coin toss.) Note this simplifies your general statement to $E(min(0, Z)) = min(E(Z)) = E(Z)$, which I hope you can see is clearly not true.
Equality holds when $X, Y$ are independent. Pick $X$ to be a coin toss with mean zero. This guarantees the RHS is zero. Can you think of a $Y$ for which the LHS is nonzero? (You can choose $Y$ to be a function of $X$.)
If $X$ is discrete and nonnegative and takes value zero with positive probability, what values does $1/X$ take?
answered Dec 18 '18 at 5:19
RileyRiley
1825
$endgroup$
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
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A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
add a comment |
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1 Answer
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1 Answer
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$begingroup$
To see that they are false, all you need to do is provide a counterexample for each, ideally one for which the LHS and RHS can be easily computed.
Here are some suggestions:
Take $X, Y$ degenerate and both equal to zero with probability one. Can you think of a $Z$ which has expected value zero (so the RHS is zero) but for which the LHS is negative? (You can choose a coin toss.) Note this simplifies your general statement to $E(min(0, Z)) = min(E(Z)) = E(Z)$, which I hope you can see is clearly not true.
Equality holds when $X, Y$ are independent. Pick $X$ to be a coin toss with mean zero. This guarantees the RHS is zero. Can you think of a $Y$ for which the LHS is nonzero? (You can choose $Y$ to be a function of $X$.)
If $X$ is discrete and nonnegative and takes value zero with positive probability, what values does $1/X$ take?
answered Dec 18 '18 at 5:19
RileyRiley
1825
$endgroup$
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
$begingroup$
A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
add a comment |
$begingroup$
To see that they are false, all you need to do is provide a counterexample for each, ideally one for which the LHS and RHS can be easily computed.
Here are some suggestions:
Take $X, Y$ degenerate and both equal to zero with probability one. Can you think of a $Z$ which has expected value zero (so the RHS is zero) but for which the LHS is negative? (You can choose a coin toss.) Note this simplifies your general statement to $E(min(0, Z)) = min(E(Z)) = E(Z)$, which I hope you can see is clearly not true.
Equality holds when $X, Y$ are independent. Pick $X$ to be a coin toss with mean zero. This guarantees the RHS is zero. Can you think of a $Y$ for which the LHS is nonzero? (You can choose $Y$ to be a function of $X$.)
If $X$ is discrete and nonnegative and takes value zero with positive probability, what values does $1/X$ take?
answered Dec 18 '18 at 5:19
RileyRiley
1825
$endgroup$
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
$begingroup$
A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
add a comment |
$begingroup$
To see that they are false, all you need to do is provide a counterexample for each, ideally one for which the LHS and RHS can be easily computed.
Here are some suggestions:
Take $X, Y$ degenerate and both equal to zero with probability one. Can you think of a $Z$ which has expected value zero (so the RHS is zero) but for which the LHS is negative? (You can choose a coin toss.) Note this simplifies your general statement to $E(min(0, Z)) = min(E(Z)) = E(Z)$, which I hope you can see is clearly not true.
Equality holds when $X, Y$ are independent. Pick $X$ to be a coin toss with mean zero. This guarantees the RHS is zero. Can you think of a $Y$ for which the LHS is nonzero? (You can choose $Y$ to be a function of $X$.)
If $X$ is discrete and nonnegative and takes value zero with positive probability, what values does $1/X$ take?
answered Dec 18 '18 at 5:19
RileyRiley
1825
$endgroup$
To see that they are false, all you need to do is provide a counterexample for each, ideally one for which the LHS and RHS can be easily computed.
Here are some suggestions:
Take $X, Y$ degenerate and both equal to zero with probability one. Can you think of a $Z$ which has expected value zero (so the RHS is zero) but for which the LHS is negative? (You can choose a coin toss.) Note this simplifies your general statement to $E(min(0, Z)) = min(E(Z)) = E(Z)$, which I hope you can see is clearly not true.
Equality holds when $X, Y$ are independent. Pick $X$ to be a coin toss with mean zero. This guarantees the RHS is zero. Can you think of a $Y$ for which the LHS is nonzero? (You can choose $Y$ to be a function of $X$.)
If $X$ is discrete and nonnegative and takes value zero with positive probability, what values does $1/X$ take?
answered Dec 18 '18 at 5:19
RileyRiley
1825
answered Dec 18 '18 at 5:19
RileyRiley
1825
answered Dec 18 '18 at 5:19
RileyRiley
1825
answered Dec 18 '18 at 5:19
RileyRiley
1825
1825
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
$begingroup$
A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
add a comment |
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
$begingroup$
A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
$begingroup$
How would these all 3 cases be translated to coin tosses scenarios to prove that they are false?
$endgroup$
– Toby
Dec 18 '18 at 5:28
$begingroup$
A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
$begingroup$
A coin toss can be modelled as a random variable taking 2 values, "heads" and "tails", with equal probability. For 1: If $X, Y$ are degenerate, this corresponds to the case where both heads and tails values are the same. $Z$ can be taken as giving "heads" a positive value and "tails" a negative value. (What that value is, I leave to you.) In other words, I think of a coin toss as a random variable $X$ which takes two values $a, b in mathbb{R}$ with equal probability.
$endgroup$
– Riley
Dec 18 '18 at 7:02
add a comment |
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$begingroup$
For 3: Just take an example coin toss scenario and compute those values. There are only two choices for $X$ (0 or 1). Or make it 1 or 2 if you prefer. All you need is one particular example to show the equality does not hold. It would be unfortunate if you end up not trying any examples at all.
$endgroup$
– Michael
Dec 18 '18 at 5:12