Calculate $E(Y)$ where $Y=X^{1.5}$












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why not variable change $y:=2x$?
    $endgroup$
    – Jimmy R.
    Dec 13 '18 at 1:34
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why not variable change $y:=2x$?
    $endgroup$
    – Jimmy R.
    Dec 13 '18 at 1:34














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 1:34







K Split X

















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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $Gamma(5/2) = 3pi^{1/2}/4$



$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $Gamma(5/2) = 3pi^{1/2}/4$



$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $Gamma(5/2) = 3pi^{1/2}/4$



$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $Gamma(5/2) = 3pi^{1/2}/4$



$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$






share|cite|improve this answer









$endgroup$



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 $Gamma(5/2) = 3pi^{1/2}/4$



$$
mathbb{E}[X^{3/2}] = 3cdot2^{-7/2}pi^{1/2}
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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