Difference of i.i.d. random variables is i.i.d.?
$begingroup$
Let $X_1,... , X_{500}$ be i.i.d. random variables with expected value
2 and variance 3. The random variables $Y_1,..., Y_{500}$ are independent of
the X variables, also i.i.d., but they have expected value 2 and
variance 2. Use the CLT to estimate $P(sum_{i=1}^{500}X_i> sum_{i=1}^{500}Y_i + 50)$. Hint. Use the CLT for the random variables $X_1-Y_1,X_2-Y_2, . . . .$
So following the hint I consider random variables $X_1-Y_1,X_2-Y_2, . . . ,X_{500}-Y_{500}.$ I have an idea how to solve this, using CLT is pretty straightforward, but in order to use the CLT, we need these random variables to be i.i.d, however I'm not sure about that, can somebody clarify this for me? Can we apply CLT?
probability
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
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add a comment |
$begingroup$
Let $X_1,... , X_{500}$ be i.i.d. random variables with expected value
2 and variance 3. The random variables $Y_1,..., Y_{500}$ are independent of
the X variables, also i.i.d., but they have expected value 2 and
variance 2. Use the CLT to estimate $P(sum_{i=1}^{500}X_i> sum_{i=1}^{500}Y_i + 50)$. Hint. Use the CLT for the random variables $X_1-Y_1,X_2-Y_2, . . . .$
So following the hint I consider random variables $X_1-Y_1,X_2-Y_2, . . . ,X_{500}-Y_{500}.$ I have an idea how to solve this, using CLT is pretty straightforward, but in order to use the CLT, we need these random variables to be i.i.d, however I'm not sure about that, can somebody clarify this for me? Can we apply CLT?
probability
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
$endgroup$
add a comment |
$begingroup$
Let $X_1,... , X_{500}$ be i.i.d. random variables with expected value
2 and variance 3. The random variables $Y_1,..., Y_{500}$ are independent of
the X variables, also i.i.d., but they have expected value 2 and
variance 2. Use the CLT to estimate $P(sum_{i=1}^{500}X_i> sum_{i=1}^{500}Y_i + 50)$. Hint. Use the CLT for the random variables $X_1-Y_1,X_2-Y_2, . . . .$
So following the hint I consider random variables $X_1-Y_1,X_2-Y_2, . . . ,X_{500}-Y_{500}.$ I have an idea how to solve this, using CLT is pretty straightforward, but in order to use the CLT, we need these random variables to be i.i.d, however I'm not sure about that, can somebody clarify this for me? Can we apply CLT?
probability
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
$endgroup$
Let $X_1,... , X_{500}$ be i.i.d. random variables with expected value
2 and variance 3. The random variables $Y_1,..., Y_{500}$ are independent of
the X variables, also i.i.d., but they have expected value 2 and
variance 2. Use the CLT to estimate $P(sum_{i=1}^{500}X_i> sum_{i=1}^{500}Y_i + 50)$. Hint. Use the CLT for the random variables $X_1-Y_1,X_2-Y_2, . . . .$
So following the hint I consider random variables $X_1-Y_1,X_2-Y_2, . . . ,X_{500}-Y_{500}.$ I have an idea how to solve this, using CLT is pretty straightforward, but in order to use the CLT, we need these random variables to be i.i.d, however I'm not sure about that, can somebody clarify this for me? Can we apply CLT?
probability
probability
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
asked Dec 9 '18 at 16:57
dxdydzdxdydz
1869
1869
add a comment |
add a comment |
1 Answer
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Yes, it's fine to use the CLT in this way.
It's true, though perhaps not quite trivial, that if ${X_i}$ is an iid collection of variables that is independent of ${Y_i}$, another iid collection of variables, then the collection ${X_i - Y_i}$ is an iid collection of variables as well. Independence isn't terribly hard to see; in this problem setup, every subset of $X_i$ and $Y_i$ variables is described to be independent of any other subset of $X_i$ and $Y_i$ variables so long as the subsets don't overlap.
For identical distribution, let's suppose the variables are all continuous and have density functions $f(x)$ for ${X_i}$ and $g(x)$ for ${Y_i}$. The joint density of the pair $(X_1, Y_1)$ would therefore be $f(x)g(x)$ by independence; if you wanted the density of $X_1 - Y_1$, you would do the standard convolution techniques. Nothing would change for the density of $X_2 - Y_2$, or for $X_3 - Y_3$, etc. And if the variables are discrete instead of continuous, then replace "density function" above with "probability function" throughout.
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
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$begingroup$
Yes, it's fine to use the CLT in this way.
It's true, though perhaps not quite trivial, that if ${X_i}$ is an iid collection of variables that is independent of ${Y_i}$, another iid collection of variables, then the collection ${X_i - Y_i}$ is an iid collection of variables as well. Independence isn't terribly hard to see; in this problem setup, every subset of $X_i$ and $Y_i$ variables is described to be independent of any other subset of $X_i$ and $Y_i$ variables so long as the subsets don't overlap.
For identical distribution, let's suppose the variables are all continuous and have density functions $f(x)$ for ${X_i}$ and $g(x)$ for ${Y_i}$. The joint density of the pair $(X_1, Y_1)$ would therefore be $f(x)g(x)$ by independence; if you wanted the density of $X_1 - Y_1$, you would do the standard convolution techniques. Nothing would change for the density of $X_2 - Y_2$, or for $X_3 - Y_3$, etc. And if the variables are discrete instead of continuous, then replace "density function" above with "probability function" throughout.
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
add a comment |
$begingroup$
Yes, it's fine to use the CLT in this way.
It's true, though perhaps not quite trivial, that if ${X_i}$ is an iid collection of variables that is independent of ${Y_i}$, another iid collection of variables, then the collection ${X_i - Y_i}$ is an iid collection of variables as well. Independence isn't terribly hard to see; in this problem setup, every subset of $X_i$ and $Y_i$ variables is described to be independent of any other subset of $X_i$ and $Y_i$ variables so long as the subsets don't overlap.
For identical distribution, let's suppose the variables are all continuous and have density functions $f(x)$ for ${X_i}$ and $g(x)$ for ${Y_i}$. The joint density of the pair $(X_1, Y_1)$ would therefore be $f(x)g(x)$ by independence; if you wanted the density of $X_1 - Y_1$, you would do the standard convolution techniques. Nothing would change for the density of $X_2 - Y_2$, or for $X_3 - Y_3$, etc. And if the variables are discrete instead of continuous, then replace "density function" above with "probability function" throughout.
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
add a comment |
$begingroup$
Yes, it's fine to use the CLT in this way.
It's true, though perhaps not quite trivial, that if ${X_i}$ is an iid collection of variables that is independent of ${Y_i}$, another iid collection of variables, then the collection ${X_i - Y_i}$ is an iid collection of variables as well. Independence isn't terribly hard to see; in this problem setup, every subset of $X_i$ and $Y_i$ variables is described to be independent of any other subset of $X_i$ and $Y_i$ variables so long as the subsets don't overlap.
For identical distribution, let's suppose the variables are all continuous and have density functions $f(x)$ for ${X_i}$ and $g(x)$ for ${Y_i}$. The joint density of the pair $(X_1, Y_1)$ would therefore be $f(x)g(x)$ by independence; if you wanted the density of $X_1 - Y_1$, you would do the standard convolution techniques. Nothing would change for the density of $X_2 - Y_2$, or for $X_3 - Y_3$, etc. And if the variables are discrete instead of continuous, then replace "density function" above with "probability function" throughout.
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
$endgroup$
Yes, it's fine to use the CLT in this way.
It's true, though perhaps not quite trivial, that if ${X_i}$ is an iid collection of variables that is independent of ${Y_i}$, another iid collection of variables, then the collection ${X_i - Y_i}$ is an iid collection of variables as well. Independence isn't terribly hard to see; in this problem setup, every subset of $X_i$ and $Y_i$ variables is described to be independent of any other subset of $X_i$ and $Y_i$ variables so long as the subsets don't overlap.
For identical distribution, let's suppose the variables are all continuous and have density functions $f(x)$ for ${X_i}$ and $g(x)$ for ${Y_i}$. The joint density of the pair $(X_1, Y_1)$ would therefore be $f(x)g(x)$ by independence; if you wanted the density of $X_1 - Y_1$, you would do the standard convolution techniques. Nothing would change for the density of $X_2 - Y_2$, or for $X_3 - Y_3$, etc. And if the variables are discrete instead of continuous, then replace "density function" above with "probability function" throughout.
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
answered Dec 9 '18 at 17:52
Aaron MontgomeryAaron Montgomery
4,732523
4,732523
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