Show that $E(Y^p)=int_{0}^{infty}px^{p-1}P(Ygeq x)dx$












1












$begingroup$


Studying some probability theory and I came across this question. Show that if $Y$ is a non-negative random variable, and $p>0$, $$E(Y^p)=int_{0}^{infty}px^{p-1}P(Ygeq x)dx$$
I'm a bit stuck on this question: I know that by Markov's inequality, $frac{E(Y^p)}{y^p}geq P(Y>y)$ since $Y=lvert Yrvert$ here. Also, $y^p=int_{0}^y px^{p-1}dx$ and I believe these two facts can be used to solve the problem, but I'm not sure how exactly.



Could anyone give me some help, point me in the right direction?






real-analysis probability







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  • $begingroup$
    Can $Y$ take on negative values?
    $endgroup$
    – Frpzzd
    Oct 29 '18 at 23:11










  • $begingroup$
    @Frpzzd $Y$ is non-negative.
    $endgroup$
    – Mog
    Oct 29 '18 at 23:11










  • $begingroup$
    math.stackexchange.com/questions/63756/tail-sum-for-expectation + change of variables
    $endgroup$
    – d.k.o.
    Oct 29 '18 at 23:17












  • $begingroup$
    This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post.
    $endgroup$
    – Lee David Chung Lin
    Nov 13 '18 at 12:47
















1












$begingroup$


Studying some probability theory and I came across this question. Show that if $Y$ is a non-negative random variable, and $p>0$, $$E(Y^p)=int_{0}^{infty}px^{p-1}P(Ygeq x)dx$$
I'm a bit stuck on this question: I know that by Markov's inequality, $frac{E(Y^p)}{y^p}geq P(Y>y)$ since $Y=lvert Yrvert$ here. Also, $y^p=int_{0}^y px^{p-1}dx$ and I believe these two facts can be used to solve the problem, but I'm not sure how exactly.



Could anyone give me some help, point me in the right direction?






real-analysis probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can $Y$ take on negative values?
    $endgroup$
    – Frpzzd
    Oct 29 '18 at 23:11










  • $begingroup$
    @Frpzzd $Y$ is non-negative.
    $endgroup$
    – Mog
    Oct 29 '18 at 23:11










  • $begingroup$
    math.stackexchange.com/questions/63756/tail-sum-for-expectation + change of variables
    $endgroup$
    – d.k.o.
    Oct 29 '18 at 23:17












  • $begingroup$
    This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post.
    $endgroup$
    – Lee David Chung Lin
    Nov 13 '18 at 12:47














1












1








1


2



$begingroup$


Studying some probability theory and I came across this question. Show that if $Y$ is a non-negative random variable, and $p>0$, $$E(Y^p)=int_{0}^{infty}px^{p-1}P(Ygeq x)dx$$
I'm a bit stuck on this question: I know that by Markov's inequality, $frac{E(Y^p)}{y^p}geq P(Y>y)$ since $Y=lvert Yrvert$ here. Also, $y^p=int_{0}^y px^{p-1}dx$ and I believe these two facts can be used to solve the problem, but I'm not sure how exactly.



Could anyone give me some help, point me in the right direction?






real-analysis probability







share|cite|improve this question











$endgroup$




Studying some probability theory and I came across this question. Show that if $Y$ is a non-negative random variable, and $p>0$, $$E(Y^p)=int_{0}^{infty}px^{p-1}P(Ygeq x)dx$$
I'm a bit stuck on this question: I know that by Markov's inequality, $frac{E(Y^p)}{y^p}geq P(Y>y)$ since $Y=lvert Yrvert$ here. Also, $y^p=int_{0}^y px^{p-1}dx$ and I believe these two facts can be used to solve the problem, but I'm not sure how exactly.



Could anyone give me some help, point me in the right direction?






real-analysis probability




real-analysis probability






share|cite|improve this question















share|cite|improve this question













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edited Dec 12 '18 at 4:06







Mog

















asked Oct 29 '18 at 23:09









MogMog

549




549












  • $begingroup$
    Can $Y$ take on negative values?
    $endgroup$
    – Frpzzd
    Oct 29 '18 at 23:11










  • $begingroup$
    @Frpzzd $Y$ is non-negative.
    $endgroup$
    – Mog
    Oct 29 '18 at 23:11










  • $begingroup$
    math.stackexchange.com/questions/63756/tail-sum-for-expectation + change of variables
    $endgroup$
    – d.k.o.
    Oct 29 '18 at 23:17












  • $begingroup$
    This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post.
    $endgroup$
    – Lee David Chung Lin
    Nov 13 '18 at 12:47


















  • $begingroup$
    Can $Y$ take on negative values?
    $endgroup$
    – Frpzzd
    Oct 29 '18 at 23:11










  • $begingroup$
    @Frpzzd $Y$ is non-negative.
    $endgroup$
    – Mog
    Oct 29 '18 at 23:11










  • $begingroup$
    math.stackexchange.com/questions/63756/tail-sum-for-expectation + change of variables
    $endgroup$
    – d.k.o.
    Oct 29 '18 at 23:17












  • $begingroup$
    This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post.
    $endgroup$
    – Lee David Chung Lin
    Nov 13 '18 at 12:47
















$begingroup$
Can $Y$ take on negative values?
$endgroup$
– Frpzzd
Oct 29 '18 at 23:11




$begingroup$
Can $Y$ take on negative values?
$endgroup$
– Frpzzd
Oct 29 '18 at 23:11












$begingroup$
@Frpzzd $Y$ is non-negative.
$endgroup$
– Mog
Oct 29 '18 at 23:11




$begingroup$
@Frpzzd $Y$ is non-negative.
$endgroup$
– Mog
Oct 29 '18 at 23:11












$begingroup$
math.stackexchange.com/questions/63756/tail-sum-for-expectation + change of variables
$endgroup$
– d.k.o.
Oct 29 '18 at 23:17






$begingroup$
math.stackexchange.com/questions/63756/tail-sum-for-expectation + change of variables
$endgroup$
– d.k.o.
Oct 29 '18 at 23:17














$begingroup$
This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post.
$endgroup$
– Lee David Chung Lin
Nov 13 '18 at 12:47




$begingroup$
This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post.
$endgroup$
– Lee David Chung Lin
Nov 13 '18 at 12:47










3 Answers
3






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oldest

votes


















2












$begingroup$

If you are willing to use Tonelli's theorem (as opposed to something more elementary) you have
$$int_0^infty px^{p-1} P(Y ge x) , dx = int_0^infty px^{p-1} int_Omega mathbb{1}_{{Y ge x}} , dP dx = int_Omega int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dxdP.$$
The inner integral evaluates as
$$int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dx = int_0^Y px^{p-1} , dx = Y^p.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
    $endgroup$
    – Mog
    Oct 29 '18 at 23:20












  • $begingroup$
    Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
    $endgroup$
    – Umberto P.
    Oct 29 '18 at 23:23










  • $begingroup$
    On any measure space $mu(E) = int mathbb{1}_E , dmu.$
    $endgroup$
    – Umberto P.
    Oct 29 '18 at 23:24












  • $begingroup$
    I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
    $endgroup$
    – Mog
    Oct 29 '18 at 23:38





















1












$begingroup$

If $Y$ has density $f(y)$, the trick is to (1) put $y^p=int_0^y px^{p-1},dx$ in the formula for $E(Y^p)$, obtaining a double integral,
then (2) interchange the order of integration:
$$begin{align}
E(Y^p) &= int_{y=0}^infty y^pf(y),dy\
&stackrel{(1)}=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}right)f(y),dy\
&=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}f(y)right),dy\
&stackrel{(2)}=int_{x=0}^inftyleft(int_{y=x}^infty px^{p-1}f(y)right),dx\
&=int_{x=0}^infty px^{p-1}left(int_{y=x}^infty f(y)right),dx\
&=int_{x=0}^infty px^{p-1} P(Yge x),dx
end{align}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
    $endgroup$
    – Mog
    Oct 30 '18 at 0:09










  • $begingroup$
    @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
    $endgroup$
    – grand_chat
    Oct 30 '18 at 0:28



















0












$begingroup$

begin{align*} E [ Y^p] & = E [int_0^{Y^p} ds ]\
& = E [ int_0^infty {bf 1}_{[0,Y^p)}(s) ds ] \
& = int_0^infty P(s< Y^p) ds \
& = int_0^infty P(Y > s^{1/p} ds \
& underset{x=s^{1/p}}{=} int_0^infty P(Y>x) p x^{p-1} dx.
end{align*}






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If you are willing to use Tonelli's theorem (as opposed to something more elementary) you have
    $$int_0^infty px^{p-1} P(Y ge x) , dx = int_0^infty px^{p-1} int_Omega mathbb{1}_{{Y ge x}} , dP dx = int_Omega int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dxdP.$$
    The inner integral evaluates as
    $$int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dx = int_0^Y px^{p-1} , dx = Y^p.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
      $endgroup$
      – Mog
      Oct 29 '18 at 23:20












    • $begingroup$
      Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:23










    • $begingroup$
      On any measure space $mu(E) = int mathbb{1}_E , dmu.$
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:24












    • $begingroup$
      I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
      $endgroup$
      – Mog
      Oct 29 '18 at 23:38


















    2












    $begingroup$

    If you are willing to use Tonelli's theorem (as opposed to something more elementary) you have
    $$int_0^infty px^{p-1} P(Y ge x) , dx = int_0^infty px^{p-1} int_Omega mathbb{1}_{{Y ge x}} , dP dx = int_Omega int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dxdP.$$
    The inner integral evaluates as
    $$int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dx = int_0^Y px^{p-1} , dx = Y^p.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
      $endgroup$
      – Mog
      Oct 29 '18 at 23:20












    • $begingroup$
      Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:23










    • $begingroup$
      On any measure space $mu(E) = int mathbb{1}_E , dmu.$
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:24












    • $begingroup$
      I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
      $endgroup$
      – Mog
      Oct 29 '18 at 23:38
















    2












    2








    2





    $begingroup$

    If you are willing to use Tonelli's theorem (as opposed to something more elementary) you have
    $$int_0^infty px^{p-1} P(Y ge x) , dx = int_0^infty px^{p-1} int_Omega mathbb{1}_{{Y ge x}} , dP dx = int_Omega int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dxdP.$$
    The inner integral evaluates as
    $$int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dx = int_0^Y px^{p-1} , dx = Y^p.$$






    share|cite|improve this answer









    $endgroup$



    If you are willing to use Tonelli's theorem (as opposed to something more elementary) you have
    $$int_0^infty px^{p-1} P(Y ge x) , dx = int_0^infty px^{p-1} int_Omega mathbb{1}_{{Y ge x}} , dP dx = int_Omega int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dxdP.$$
    The inner integral evaluates as
    $$int_0^infty px^{p-1} mathbb{1}_{{Y ge x}} , dx = int_0^Y px^{p-1} , dx = Y^p.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 29 '18 at 23:18









    Umberto P.Umberto P.

    40.3k13370




    40.3k13370












    • $begingroup$
      Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
      $endgroup$
      – Mog
      Oct 29 '18 at 23:20












    • $begingroup$
      Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:23










    • $begingroup$
      On any measure space $mu(E) = int mathbb{1}_E , dmu.$
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:24












    • $begingroup$
      I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
      $endgroup$
      – Mog
      Oct 29 '18 at 23:38




















    • $begingroup$
      Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
      $endgroup$
      – Mog
      Oct 29 '18 at 23:20












    • $begingroup$
      Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:23










    • $begingroup$
      On any measure space $mu(E) = int mathbb{1}_E , dmu.$
      $endgroup$
      – Umberto P.
      Oct 29 '18 at 23:24












    • $begingroup$
      I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
      $endgroup$
      – Mog
      Oct 29 '18 at 23:38


















    $begingroup$
    Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
    $endgroup$
    – Mog
    Oct 29 '18 at 23:20






    $begingroup$
    Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived?
    $endgroup$
    – Mog
    Oct 29 '18 at 23:20














    $begingroup$
    Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
    $endgroup$
    – Umberto P.
    Oct 29 '18 at 23:23




    $begingroup$
    Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter.
    $endgroup$
    – Umberto P.
    Oct 29 '18 at 23:23












    $begingroup$
    On any measure space $mu(E) = int mathbb{1}_E , dmu.$
    $endgroup$
    – Umberto P.
    Oct 29 '18 at 23:24






    $begingroup$
    On any measure space $mu(E) = int mathbb{1}_E , dmu.$
    $endgroup$
    – Umberto P.
    Oct 29 '18 at 23:24














    $begingroup$
    I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
    $endgroup$
    – Mog
    Oct 29 '18 at 23:38






    $begingroup$
    I believe I can apply Fubini's by stating that $px^{p-1}mathbb 1_{Ygeq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.)
    $endgroup$
    – Mog
    Oct 29 '18 at 23:38













    1












    $begingroup$

    If $Y$ has density $f(y)$, the trick is to (1) put $y^p=int_0^y px^{p-1},dx$ in the formula for $E(Y^p)$, obtaining a double integral,
    then (2) interchange the order of integration:
    $$begin{align}
    E(Y^p) &= int_{y=0}^infty y^pf(y),dy\
    &stackrel{(1)}=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}right)f(y),dy\
    &=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}f(y)right),dy\
    &stackrel{(2)}=int_{x=0}^inftyleft(int_{y=x}^infty px^{p-1}f(y)right),dx\
    &=int_{x=0}^infty px^{p-1}left(int_{y=x}^infty f(y)right),dx\
    &=int_{x=0}^infty px^{p-1} P(Yge x),dx
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
      $endgroup$
      – Mog
      Oct 30 '18 at 0:09










    • $begingroup$
      @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
      $endgroup$
      – grand_chat
      Oct 30 '18 at 0:28
















    1












    $begingroup$

    If $Y$ has density $f(y)$, the trick is to (1) put $y^p=int_0^y px^{p-1},dx$ in the formula for $E(Y^p)$, obtaining a double integral,
    then (2) interchange the order of integration:
    $$begin{align}
    E(Y^p) &= int_{y=0}^infty y^pf(y),dy\
    &stackrel{(1)}=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}right)f(y),dy\
    &=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}f(y)right),dy\
    &stackrel{(2)}=int_{x=0}^inftyleft(int_{y=x}^infty px^{p-1}f(y)right),dx\
    &=int_{x=0}^infty px^{p-1}left(int_{y=x}^infty f(y)right),dx\
    &=int_{x=0}^infty px^{p-1} P(Yge x),dx
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
      $endgroup$
      – Mog
      Oct 30 '18 at 0:09










    • $begingroup$
      @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
      $endgroup$
      – grand_chat
      Oct 30 '18 at 0:28














    1












    1








    1





    $begingroup$

    If $Y$ has density $f(y)$, the trick is to (1) put $y^p=int_0^y px^{p-1},dx$ in the formula for $E(Y^p)$, obtaining a double integral,
    then (2) interchange the order of integration:
    $$begin{align}
    E(Y^p) &= int_{y=0}^infty y^pf(y),dy\
    &stackrel{(1)}=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}right)f(y),dy\
    &=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}f(y)right),dy\
    &stackrel{(2)}=int_{x=0}^inftyleft(int_{y=x}^infty px^{p-1}f(y)right),dx\
    &=int_{x=0}^infty px^{p-1}left(int_{y=x}^infty f(y)right),dx\
    &=int_{x=0}^infty px^{p-1} P(Yge x),dx
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$



    If $Y$ has density $f(y)$, the trick is to (1) put $y^p=int_0^y px^{p-1},dx$ in the formula for $E(Y^p)$, obtaining a double integral,
    then (2) interchange the order of integration:
    $$begin{align}
    E(Y^p) &= int_{y=0}^infty y^pf(y),dy\
    &stackrel{(1)}=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}right)f(y),dy\
    &=int_{y=0}^inftyleft(int_{x=0}^y px^{p-1}f(y)right),dy\
    &stackrel{(2)}=int_{x=0}^inftyleft(int_{y=x}^infty px^{p-1}f(y)right),dx\
    &=int_{x=0}^infty px^{p-1}left(int_{y=x}^infty f(y)right),dx\
    &=int_{x=0}^infty px^{p-1} P(Yge x),dx
    end{align}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 29 '18 at 23:29









    grand_chatgrand_chat

    20.5k11327




    20.5k11327












    • $begingroup$
      Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
      $endgroup$
      – Mog
      Oct 30 '18 at 0:09










    • $begingroup$
      @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
      $endgroup$
      – grand_chat
      Oct 30 '18 at 0:28


















    • $begingroup$
      Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
      $endgroup$
      – Mog
      Oct 30 '18 at 0:09










    • $begingroup$
      @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
      $endgroup$
      – grand_chat
      Oct 30 '18 at 0:28
















    $begingroup$
    Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
    $endgroup$
    – Mog
    Oct 30 '18 at 0:09




    $begingroup$
    Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right?
    $endgroup$
    – Mog
    Oct 30 '18 at 0:09












    $begingroup$
    @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
    $endgroup$
    – grand_chat
    Oct 30 '18 at 0:28




    $begingroup$
    @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do.
    $endgroup$
    – grand_chat
    Oct 30 '18 at 0:28











    0












    $begingroup$

    begin{align*} E [ Y^p] & = E [int_0^{Y^p} ds ]\
    & = E [ int_0^infty {bf 1}_{[0,Y^p)}(s) ds ] \
    & = int_0^infty P(s< Y^p) ds \
    & = int_0^infty P(Y > s^{1/p} ds \
    & underset{x=s^{1/p}}{=} int_0^infty P(Y>x) p x^{p-1} dx.
    end{align*}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      begin{align*} E [ Y^p] & = E [int_0^{Y^p} ds ]\
      & = E [ int_0^infty {bf 1}_{[0,Y^p)}(s) ds ] \
      & = int_0^infty P(s< Y^p) ds \
      & = int_0^infty P(Y > s^{1/p} ds \
      & underset{x=s^{1/p}}{=} int_0^infty P(Y>x) p x^{p-1} dx.
      end{align*}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        begin{align*} E [ Y^p] & = E [int_0^{Y^p} ds ]\
        & = E [ int_0^infty {bf 1}_{[0,Y^p)}(s) ds ] \
        & = int_0^infty P(s< Y^p) ds \
        & = int_0^infty P(Y > s^{1/p} ds \
        & underset{x=s^{1/p}}{=} int_0^infty P(Y>x) p x^{p-1} dx.
        end{align*}






        share|cite|improve this answer









        $endgroup$



        begin{align*} E [ Y^p] & = E [int_0^{Y^p} ds ]\
        & = E [ int_0^infty {bf 1}_{[0,Y^p)}(s) ds ] \
        & = int_0^infty P(s< Y^p) ds \
        & = int_0^infty P(Y > s^{1/p} ds \
        & underset{x=s^{1/p}}{=} int_0^infty P(Y>x) p x^{p-1} dx.
        end{align*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 13 '18 at 13:01









        FnacoolFnacool

        5,291611




        5,291611






























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