Calculate $E(Y)$ where $Y=X^{1.5}$
$begingroup$
Let $X$ be a rv which is $Exp(lambda=2)$.
The pdf of $X$ is given by $f_X(x)=2e^{-2x}, xgeq 0$ (and $0$ otherwise).
We define $Y=X^{1.5}$ and ask $E(Y)$.
It does not look like the moment generating function can help me here. So, instead I evaluate the integral:
$$2int^{infty}_{0}e^{-2x}x^{1.5}dx$$
I know the Gamma function is given by: $$int_{0}^{infty}x^{alpha-1}e^{-x}dx$$ but I have $e^{-2x}$ in my original integral. So I don't think this can help me either.
My question is the following:
1) If I were to evaluate the integral, it seems very messy by parts. Is there a simple way to evaluate the integral?
2) Is there a simpler way to solve the question that I am missing?
probability integration statistics probability-distributions gamma-function
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
$endgroup$
add a comment |
$begingroup$
Let $X$ be a rv which is $Exp(lambda=2)$.
The pdf of $X$ is given by $f_X(x)=2e^{-2x}, xgeq 0$ (and $0$ otherwise).
We define $Y=X^{1.5}$ and ask $E(Y)$.
It does not look like the moment generating function can help me here. So, instead I evaluate the integral:
$$2int^{infty}_{0}e^{-2x}x^{1.5}dx$$
I know the Gamma function is given by: $$int_{0}^{infty}x^{alpha-1}e^{-x}dx$$ but I have $e^{-2x}$ in my original integral. So I don't think this can help me either.
My question is the following:
1) If I were to evaluate the integral, it seems very messy by parts. Is there a simple way to evaluate the integral?
2) Is there a simpler way to solve the question that I am missing?
probability integration statistics probability-distributions gamma-function
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
$endgroup$
1
$begingroup$
Why not variable change $y:=2x$?
$endgroup$
– Jimmy R.
Dec 13 '18 at 1:34
add a comment |
$begingroup$
Let $X$ be a rv which is $Exp(lambda=2)$.
The pdf of $X$ is given by $f_X(x)=2e^{-2x}, xgeq 0$ (and $0$ otherwise).
We define $Y=X^{1.5}$ and ask $E(Y)$.
It does not look like the moment generating function can help me here. So, instead I evaluate the integral:
$$2int^{infty}_{0}e^{-2x}x^{1.5}dx$$
I know the Gamma function is given by: $$int_{0}^{infty}x^{alpha-1}e^{-x}dx$$ but I have $e^{-2x}$ in my original integral. So I don't think this can help me either.
My question is the following:
1) If I were to evaluate the integral, it seems very messy by parts. Is there a simple way to evaluate the integral?
2) Is there a simpler way to solve the question that I am missing?
probability integration statistics probability-distributions gamma-function
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
$endgroup$
Let $X$ be a rv which is $Exp(lambda=2)$.
The pdf of $X$ is given by $f_X(x)=2e^{-2x}, xgeq 0$ (and $0$ otherwise).
We define $Y=X^{1.5}$ and ask $E(Y)$.
It does not look like the moment generating function can help me here. So, instead I evaluate the integral:
$$2int^{infty}_{0}e^{-2x}x^{1.5}dx$$
I know the Gamma function is given by: $$int_{0}^{infty}x^{alpha-1}e^{-x}dx$$ but I have $e^{-2x}$ in my original integral. So I don't think this can help me either.
My question is the following:
1) If I were to evaluate the integral, it seems very messy by parts. Is there a simple way to evaluate the integral?
2) Is there a simpler way to solve the question that I am missing?
probability integration statistics probability-distributions gamma-function
probability integration statistics probability-distributions gamma-function
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
edited Dec 13 '18 at 1:34
K Split X
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
asked Dec 13 '18 at 1:32
K Split XK Split X
4,23011132
4,23011132
1
$begingroup$
Why not variable change $y:=2x$?
$endgroup$
– Jimmy R.
Dec 13 '18 at 1:34
add a comment |
1
$begingroup$
Why not variable change $y:=2x$?
$endgroup$
– Jimmy R.
Dec 13 '18 at 1:34
1
1
$begingroup$
Why not variable change $y:=2x$?
$endgroup$
– Jimmy R.
Dec 13 '18 at 1:34
$begingroup$
Why not variable change $y:=2x$?
$endgroup$
– Jimmy R.
Dec 13 '18 at 1:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just make a change of variables $u = 2x$, so that ${rm d}x = {rm d} u / 2$ and
$$
2int_0^{+infty}x^{3/2}e^{-2x}{rm d}x = 2^{-3/2} int_0^{+infty}u^{3/2}e^{-u}{rm d}u = 2^{-3/2} int_0^{+infty}u^{5/2 - 1}e^{-u}{rm d}u = 2^{-3/2}Gamma(5/2)
$$
You can further simplify this by using $Gamma(5/2) = 3pi^{1/2}/4$
$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
add a comment |
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$begingroup$
Just make a change of variables $u = 2x$, so that ${rm d}x = {rm d} u / 2$ and
$$
2int_0^{+infty}x^{3/2}e^{-2x}{rm d}x = 2^{-3/2} int_0^{+infty}u^{3/2}e^{-u}{rm d}u = 2^{-3/2} int_0^{+infty}u^{5/2 - 1}e^{-u}{rm d}u = 2^{-3/2}Gamma(5/2)
$$
You can further simplify this by using $Gamma(5/2) = 3pi^{1/2}/4$
$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
add a comment |
$begingroup$
Just make a change of variables $u = 2x$, so that ${rm d}x = {rm d} u / 2$ and
$$
2int_0^{+infty}x^{3/2}e^{-2x}{rm d}x = 2^{-3/2} int_0^{+infty}u^{3/2}e^{-u}{rm d}u = 2^{-3/2} int_0^{+infty}u^{5/2 - 1}e^{-u}{rm d}u = 2^{-3/2}Gamma(5/2)
$$
You can further simplify this by using $Gamma(5/2) = 3pi^{1/2}/4$
$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
add a comment |
$begingroup$
Just make a change of variables $u = 2x$, so that ${rm d}x = {rm d} u / 2$ and
$$
2int_0^{+infty}x^{3/2}e^{-2x}{rm d}x = 2^{-3/2} int_0^{+infty}u^{3/2}e^{-u}{rm d}u = 2^{-3/2} int_0^{+infty}u^{5/2 - 1}e^{-u}{rm d}u = 2^{-3/2}Gamma(5/2)
$$
You can further simplify this by using $Gamma(5/2) = 3pi^{1/2}/4$
$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
$endgroup$
Just make a change of variables $u = 2x$, so that ${rm d}x = {rm d} u / 2$ and
$$
2int_0^{+infty}x^{3/2}e^{-2x}{rm d}x = 2^{-3/2} int_0^{+infty}u^{3/2}e^{-u}{rm d}u = 2^{-3/2} int_0^{+infty}u^{5/2 - 1}e^{-u}{rm d}u = 2^{-3/2}Gamma(5/2)
$$
You can further simplify this by using $Gamma(5/2) = 3pi^{1/2}/4$
$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
answered Dec 13 '18 at 1:36
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
add a comment |
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
Sometimes it's the simplest things you miss. Thank you
$endgroup$
– K Split X
Dec 13 '18 at 3:09
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
On a second thought, do you know if this contradicts the monotonicity of expectation? Basically it says that if $X leq Y$, then $E[X]leq E[Y]$. We have that $Xleq X^{1.5}$, but $E[X]=1/lambda = 1/2$, so this does not hold, as $E[X^{1.5}]approx 0.47$
$endgroup$
– K Split X
Dec 13 '18 at 3:40
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
$begingroup$
@KSplitX It doesn't violate it because $X>Y$ for $X<1$, but $X<Y$ for $X>1$, so the condition $Xleq Y$ is not satisfied
$endgroup$
– caverac
Dec 13 '18 at 15:16
add a comment |
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$begingroup$
Why not variable change $y:=2x$?
$endgroup$
– Jimmy R.
Dec 13 '18 at 1:34