Compute $E(sqrt{X+Y})$ given that $X,Y$ are iid
$begingroup$
Compute $E(sqrt{X+Y})$ given that $X,Y$ are iid.
Assume $X,Y$ are iid both having an Exp($lambda=1)$ distribution.
Although this is the only information provided in the question, I know that since they are independent, then the joint distribution must be $$f_{X,Y}(X,Y)=e^{-x} e^{-y}$$
Then we must integrate:
$$int_{0}^{infty}int_{0}^{infty}e^{-x}e^{-y}(x+y)^{1/2}dxdy=int_{0}^{infty}e^{-y}int_{0}^{infty}e^{-x}(x+y)^{1/2}dxdy$$
This integral seems complicated to evaluate. I tried it out and it seems like I'll have to do parts atleast twice. Wolfram says the answer is $1.32934$ and I have $1.33$ as one of the options to this question.
I want to know if there's an easier way to evaluate this. Do I even need the integral? Or perhaps is there some trick to get the answer with this integral?
probability integration probability-distributions expected-value
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
$endgroup$
add a comment |
$begingroup$
Compute $E(sqrt{X+Y})$ given that $X,Y$ are iid.
Assume $X,Y$ are iid both having an Exp($lambda=1)$ distribution.
Although this is the only information provided in the question, I know that since they are independent, then the joint distribution must be $$f_{X,Y}(X,Y)=e^{-x} e^{-y}$$
Then we must integrate:
$$int_{0}^{infty}int_{0}^{infty}e^{-x}e^{-y}(x+y)^{1/2}dxdy=int_{0}^{infty}e^{-y}int_{0}^{infty}e^{-x}(x+y)^{1/2}dxdy$$
This integral seems complicated to evaluate. I tried it out and it seems like I'll have to do parts atleast twice. Wolfram says the answer is $1.32934$ and I have $1.33$ as one of the options to this question.
I want to know if there's an easier way to evaluate this. Do I even need the integral? Or perhaps is there some trick to get the answer with this integral?
probability integration probability-distributions expected-value
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
$endgroup$
$begingroup$
How so? There's a square root surrounding it
$endgroup$
– K Split X
Dec 17 '18 at 2:27
1
$begingroup$
One thing that might be worth committing to memory is that the sum of $n$ IID exponentials (with the same rate) is Gamma(n)-distributed. So, here the PDF for $Z=X+Y$ is $f(z) = z e^{-z}$ so the answer is $int_0^infty z^{3/2} e^{-z}dz.$ In general, it's probably a good approach to first find the PDF of $Z=X+Y$ using a standard approach and then calculate $E(sqrt{Z}).$
$endgroup$
– spaceisdarkgreen
Dec 17 '18 at 2:41
add a comment |
$begingroup$
Compute $E(sqrt{X+Y})$ given that $X,Y$ are iid.
Assume $X,Y$ are iid both having an Exp($lambda=1)$ distribution.
Although this is the only information provided in the question, I know that since they are independent, then the joint distribution must be $$f_{X,Y}(X,Y)=e^{-x} e^{-y}$$
Then we must integrate:
$$int_{0}^{infty}int_{0}^{infty}e^{-x}e^{-y}(x+y)^{1/2}dxdy=int_{0}^{infty}e^{-y}int_{0}^{infty}e^{-x}(x+y)^{1/2}dxdy$$
This integral seems complicated to evaluate. I tried it out and it seems like I'll have to do parts atleast twice. Wolfram says the answer is $1.32934$ and I have $1.33$ as one of the options to this question.
I want to know if there's an easier way to evaluate this. Do I even need the integral? Or perhaps is there some trick to get the answer with this integral?
probability integration probability-distributions expected-value
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
$endgroup$
Compute $E(sqrt{X+Y})$ given that $X,Y$ are iid.
Assume $X,Y$ are iid both having an Exp($lambda=1)$ distribution.
Although this is the only information provided in the question, I know that since they are independent, then the joint distribution must be $$f_{X,Y}(X,Y)=e^{-x} e^{-y}$$
Then we must integrate:
$$int_{0}^{infty}int_{0}^{infty}e^{-x}e^{-y}(x+y)^{1/2}dxdy=int_{0}^{infty}e^{-y}int_{0}^{infty}e^{-x}(x+y)^{1/2}dxdy$$
This integral seems complicated to evaluate. I tried it out and it seems like I'll have to do parts atleast twice. Wolfram says the answer is $1.32934$ and I have $1.33$ as one of the options to this question.
I want to know if there's an easier way to evaluate this. Do I even need the integral? Or perhaps is there some trick to get the answer with this integral?
probability integration probability-distributions expected-value
probability integration probability-distributions expected-value
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
asked Dec 17 '18 at 2:19
K Split XK Split X
4,30421233
4,30421233
$begingroup$
How so? There's a square root surrounding it
$endgroup$
– K Split X
Dec 17 '18 at 2:27
1
$begingroup$
One thing that might be worth committing to memory is that the sum of $n$ IID exponentials (with the same rate) is Gamma(n)-distributed. So, here the PDF for $Z=X+Y$ is $f(z) = z e^{-z}$ so the answer is $int_0^infty z^{3/2} e^{-z}dz.$ In general, it's probably a good approach to first find the PDF of $Z=X+Y$ using a standard approach and then calculate $E(sqrt{Z}).$
$endgroup$
– spaceisdarkgreen
Dec 17 '18 at 2:41
add a comment |
$begingroup$
How so? There's a square root surrounding it
$endgroup$
– K Split X
Dec 17 '18 at 2:27
1
$begingroup$
One thing that might be worth committing to memory is that the sum of $n$ IID exponentials (with the same rate) is Gamma(n)-distributed. So, here the PDF for $Z=X+Y$ is $f(z) = z e^{-z}$ so the answer is $int_0^infty z^{3/2} e^{-z}dz.$ In general, it's probably a good approach to first find the PDF of $Z=X+Y$ using a standard approach and then calculate $E(sqrt{Z}).$
$endgroup$
– spaceisdarkgreen
Dec 17 '18 at 2:41
$begingroup$
How so? There's a square root surrounding it
$endgroup$
– K Split X
Dec 17 '18 at 2:27
$begingroup$
How so? There's a square root surrounding it
$endgroup$
– K Split X
Dec 17 '18 at 2:27
1
1
$begingroup$
One thing that might be worth committing to memory is that the sum of $n$ IID exponentials (with the same rate) is Gamma(n)-distributed. So, here the PDF for $Z=X+Y$ is $f(z) = z e^{-z}$ so the answer is $int_0^infty z^{3/2} e^{-z}dz.$ In general, it's probably a good approach to first find the PDF of $Z=X+Y$ using a standard approach and then calculate $E(sqrt{Z}).$
$endgroup$
– spaceisdarkgreen
Dec 17 '18 at 2:41
$begingroup$
One thing that might be worth committing to memory is that the sum of $n$ IID exponentials (with the same rate) is Gamma(n)-distributed. So, here the PDF for $Z=X+Y$ is $f(z) = z e^{-z}$ so the answer is $int_0^infty z^{3/2} e^{-z}dz.$ In general, it's probably a good approach to first find the PDF of $Z=X+Y$ using a standard approach and then calculate $E(sqrt{Z}).$
$endgroup$
– spaceisdarkgreen
Dec 17 '18 at 2:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One way of getting the integral slightly faster is to see that everything is in terms of $x+y$ only.
So let $z=x+y$, which ranges from $0$ to $infty$, and then given $z$ we have $x$ ranges from $0$ to $z$. The Jacobian for this is just $1$.
Then the double integral becomes
$$
int_0^{infty}int_0^ze^{-z}z^{frac{1}{2}},mathrm{d}x,mathrm{d}z=int_0^{infty}e^{-z}z^{frac{3}{2}},mathrm{d}z=Gammaleft(frac{5}{2}right)=frac{3sqrt{pi}}{4}$$
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
$endgroup$
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
add a comment |
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$begingroup$
One way of getting the integral slightly faster is to see that everything is in terms of $x+y$ only.
So let $z=x+y$, which ranges from $0$ to $infty$, and then given $z$ we have $x$ ranges from $0$ to $z$. The Jacobian for this is just $1$.
Then the double integral becomes
$$
int_0^{infty}int_0^ze^{-z}z^{frac{1}{2}},mathrm{d}x,mathrm{d}z=int_0^{infty}e^{-z}z^{frac{3}{2}},mathrm{d}z=Gammaleft(frac{5}{2}right)=frac{3sqrt{pi}}{4}$$
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
$endgroup$
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
add a comment |
$begingroup$
One way of getting the integral slightly faster is to see that everything is in terms of $x+y$ only.
So let $z=x+y$, which ranges from $0$ to $infty$, and then given $z$ we have $x$ ranges from $0$ to $z$. The Jacobian for this is just $1$.
Then the double integral becomes
$$
int_0^{infty}int_0^ze^{-z}z^{frac{1}{2}},mathrm{d}x,mathrm{d}z=int_0^{infty}e^{-z}z^{frac{3}{2}},mathrm{d}z=Gammaleft(frac{5}{2}right)=frac{3sqrt{pi}}{4}$$
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
$endgroup$
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
add a comment |
$begingroup$
One way of getting the integral slightly faster is to see that everything is in terms of $x+y$ only.
So let $z=x+y$, which ranges from $0$ to $infty$, and then given $z$ we have $x$ ranges from $0$ to $z$. The Jacobian for this is just $1$.
Then the double integral becomes
$$
int_0^{infty}int_0^ze^{-z}z^{frac{1}{2}},mathrm{d}x,mathrm{d}z=int_0^{infty}e^{-z}z^{frac{3}{2}},mathrm{d}z=Gammaleft(frac{5}{2}right)=frac{3sqrt{pi}}{4}$$
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
$endgroup$
One way of getting the integral slightly faster is to see that everything is in terms of $x+y$ only.
So let $z=x+y$, which ranges from $0$ to $infty$, and then given $z$ we have $x$ ranges from $0$ to $z$. The Jacobian for this is just $1$.
Then the double integral becomes
$$
int_0^{infty}int_0^ze^{-z}z^{frac{1}{2}},mathrm{d}x,mathrm{d}z=int_0^{infty}e^{-z}z^{frac{3}{2}},mathrm{d}z=Gammaleft(frac{5}{2}right)=frac{3sqrt{pi}}{4}$$
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
answered Dec 17 '18 at 2:32
obscuransobscurans
1,152311
1,152311
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
add a comment |
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
So this is $u$ substitution. If we let $u=x+y$, then the inner integral goes from $$int_{0}^{infty}e^{-x}(x+y)^{1/2}dxtoint_{y}^{infty}e^{y-u}u^{1/2}du$$. I'm not sure how you get the $z$ on top. Or am I missing something?
$endgroup$
– K Split X
Dec 17 '18 at 2:44
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
The analogue of $u$ substitution in multiple integrals is to define $u(x,y)$ and $v(x,y)$ as your new coordinates. The factor to scale the integral is the Jacobian, which is the determinant of the matrix of partial derivatives. Similarly to $u$ substitution in a definite integral, you also need to re-express the limits of integration in the new coordinates. For 2D, it is best to draw a diagram and see what is going on.
$endgroup$
– obscurans
Dec 17 '18 at 2:47
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
$begingroup$
In this particular case, $z=x+y$ and $x=x$. The original region is the quadrant $xgeq0$, $ygeq0$. Within this region, $x+y$ goes from $0$ to $infty$. To see what the $x$ range is, consider fixing $z=x+y$. This is the diagonal line $y=z-x$, which fits in the quadrant as long as $0leq xleq z$, which is therefore the limits for the inner integral.
$endgroup$
– obscurans
Dec 17 '18 at 2:52
add a comment |
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$begingroup$
How so? There's a square root surrounding it
$endgroup$
– K Split X
Dec 17 '18 at 2:27
1
$begingroup$
One thing that might be worth committing to memory is that the sum of $n$ IID exponentials (with the same rate) is Gamma(n)-distributed. So, here the PDF for $Z=X+Y$ is $f(z) = z e^{-z}$ so the answer is $int_0^infty z^{3/2} e^{-z}dz.$ In general, it's probably a good approach to first find the PDF of $Z=X+Y$ using a standard approach and then calculate $E(sqrt{Z}).$
$endgroup$
– spaceisdarkgreen
Dec 17 '18 at 2:41