Is there a shortcut to computing $text{Cov}(X, Y)$ if we have $mathbb{E}[X]$, $mathbb{E}[Y]$ and...












1












$begingroup$


Is there a shortcut to computing $text{Cov}(X, Y)$ if we have $mathbb{E}[X]$, $mathbb{E}[Y]$ and $mathbb{E}[Ymid X]$?



I have the joint PDF



$$f(x, y) = begin{cases} lambda e^{-lambda x}/x & text{ if } 0 < y < x, x in (0, infty) \[1em] 0, &text{ otherwise.} end{cases} $$



I want to find $text{Cov}(X, Y)$. I have $mathbb{E}[X] = 1/lambda$, $mathbb{E}[Ymid X] = x/2$ and $mathbb{E}[Y] = 1/2lambda$.



I need $mathbb{E}[XY]$ to compute $text{Cov}(X, Y) = mathbb{E}[XY] - mathbb{E}[X]mathbb{E}[Y].$



Do I need to compute the double integral for it? or is there a shortcut?




My attempt:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} (xy) lambda e^{-lambda x}/x mathop{dy} mathop{dx}$$



$$= int_{0}^{infty} lambda e^{-lambda x} int_{0}^{x} ymathop{dy}mathop{dx} $$



$$= frac{lambda }{2}int_{0}^{infty} x^{2}e^{-lambda x} mathop{dx}$$



Then let $u = x^{2}, mathop{du} = 2x mathop{dx}$, $mathop{dv} = e^{-lambda x}, v = -e^{-lambda x}/lambda$. We have



$$int_{0}^{infty} x^{2} e^{-lambda x} mathop{dx} $$



$$= -frac{x^{2}e^{-lambda x}}{lambda} + int_{0}^{infty} frac{2xe^{-lambda x}}{lambda} mathop{dx} $$






probability probability-theory probability-distributions covariance







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Best I can do: $$mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY mid X]] = mathbb{E}[Xmathbb{E}[Y mid X]] = mathbb{E}[X^2/2] = dfrac{1}{2}mathbb{E}[X^2]$$
    $endgroup$
    – Clarinetist
    Dec 4 '18 at 20:56










  • $begingroup$
    that will work for me. i can get the variance because i know $X sim text{Exp}(lambda)$. then I can do $mathbb{E}[X^{2}] = (text{Var}(X) + mathbb{E}[X]^{2})$
    $endgroup$
    – joseph
    Dec 4 '18 at 20:58








  • 2




    $begingroup$
    The next step is to simplify even further the computations, noting that $$mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(XE(Ymid X))-E(X)E(E(Ymid X))$$ that is, $$mathrm{Cov}(X,Y)=E(X(tfrac12X))-E(X)E(tfrac12X)=tfrac12mathrm{var}(X)$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:17


















1












$begingroup$


Is there a shortcut to computing $text{Cov}(X, Y)$ if we have $mathbb{E}[X]$, $mathbb{E}[Y]$ and $mathbb{E}[Ymid X]$?



I have the joint PDF



$$f(x, y) = begin{cases} lambda e^{-lambda x}/x & text{ if } 0 < y < x, x in (0, infty) \[1em] 0, &text{ otherwise.} end{cases} $$



I want to find $text{Cov}(X, Y)$. I have $mathbb{E}[X] = 1/lambda$, $mathbb{E}[Ymid X] = x/2$ and $mathbb{E}[Y] = 1/2lambda$.



I need $mathbb{E}[XY]$ to compute $text{Cov}(X, Y) = mathbb{E}[XY] - mathbb{E}[X]mathbb{E}[Y].$



Do I need to compute the double integral for it? or is there a shortcut?




My attempt:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} (xy) lambda e^{-lambda x}/x mathop{dy} mathop{dx}$$



$$= int_{0}^{infty} lambda e^{-lambda x} int_{0}^{x} ymathop{dy}mathop{dx} $$



$$= frac{lambda }{2}int_{0}^{infty} x^{2}e^{-lambda x} mathop{dx}$$



Then let $u = x^{2}, mathop{du} = 2x mathop{dx}$, $mathop{dv} = e^{-lambda x}, v = -e^{-lambda x}/lambda$. We have



$$int_{0}^{infty} x^{2} e^{-lambda x} mathop{dx} $$



$$= -frac{x^{2}e^{-lambda x}}{lambda} + int_{0}^{infty} frac{2xe^{-lambda x}}{lambda} mathop{dx} $$






probability probability-theory probability-distributions covariance







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Best I can do: $$mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY mid X]] = mathbb{E}[Xmathbb{E}[Y mid X]] = mathbb{E}[X^2/2] = dfrac{1}{2}mathbb{E}[X^2]$$
    $endgroup$
    – Clarinetist
    Dec 4 '18 at 20:56










  • $begingroup$
    that will work for me. i can get the variance because i know $X sim text{Exp}(lambda)$. then I can do $mathbb{E}[X^{2}] = (text{Var}(X) + mathbb{E}[X]^{2})$
    $endgroup$
    – joseph
    Dec 4 '18 at 20:58








  • 2




    $begingroup$
    The next step is to simplify even further the computations, noting that $$mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(XE(Ymid X))-E(X)E(E(Ymid X))$$ that is, $$mathrm{Cov}(X,Y)=E(X(tfrac12X))-E(X)E(tfrac12X)=tfrac12mathrm{var}(X)$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:17
















1












1








1





$begingroup$


Is there a shortcut to computing $text{Cov}(X, Y)$ if we have $mathbb{E}[X]$, $mathbb{E}[Y]$ and $mathbb{E}[Ymid X]$?



I have the joint PDF



$$f(x, y) = begin{cases} lambda e^{-lambda x}/x & text{ if } 0 < y < x, x in (0, infty) \[1em] 0, &text{ otherwise.} end{cases} $$



I want to find $text{Cov}(X, Y)$. I have $mathbb{E}[X] = 1/lambda$, $mathbb{E}[Ymid X] = x/2$ and $mathbb{E}[Y] = 1/2lambda$.



I need $mathbb{E}[XY]$ to compute $text{Cov}(X, Y) = mathbb{E}[XY] - mathbb{E}[X]mathbb{E}[Y].$



Do I need to compute the double integral for it? or is there a shortcut?




My attempt:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} (xy) lambda e^{-lambda x}/x mathop{dy} mathop{dx}$$



$$= int_{0}^{infty} lambda e^{-lambda x} int_{0}^{x} ymathop{dy}mathop{dx} $$



$$= frac{lambda }{2}int_{0}^{infty} x^{2}e^{-lambda x} mathop{dx}$$



Then let $u = x^{2}, mathop{du} = 2x mathop{dx}$, $mathop{dv} = e^{-lambda x}, v = -e^{-lambda x}/lambda$. We have



$$int_{0}^{infty} x^{2} e^{-lambda x} mathop{dx} $$



$$= -frac{x^{2}e^{-lambda x}}{lambda} + int_{0}^{infty} frac{2xe^{-lambda x}}{lambda} mathop{dx} $$






probability probability-theory probability-distributions covariance







share|cite|improve this question









$endgroup$




Is there a shortcut to computing $text{Cov}(X, Y)$ if we have $mathbb{E}[X]$, $mathbb{E}[Y]$ and $mathbb{E}[Ymid X]$?



I have the joint PDF



$$f(x, y) = begin{cases} lambda e^{-lambda x}/x & text{ if } 0 < y < x, x in (0, infty) \[1em] 0, &text{ otherwise.} end{cases} $$



I want to find $text{Cov}(X, Y)$. I have $mathbb{E}[X] = 1/lambda$, $mathbb{E}[Ymid X] = x/2$ and $mathbb{E}[Y] = 1/2lambda$.



I need $mathbb{E}[XY]$ to compute $text{Cov}(X, Y) = mathbb{E}[XY] - mathbb{E}[X]mathbb{E}[Y].$



Do I need to compute the double integral for it? or is there a shortcut?




My attempt:



$$mathbb{E}[XY] = int_{0}^{infty} int_{0}^{x} (xy) lambda e^{-lambda x}/x mathop{dy} mathop{dx}$$



$$= int_{0}^{infty} lambda e^{-lambda x} int_{0}^{x} ymathop{dy}mathop{dx} $$



$$= frac{lambda }{2}int_{0}^{infty} x^{2}e^{-lambda x} mathop{dx}$$



Then let $u = x^{2}, mathop{du} = 2x mathop{dx}$, $mathop{dv} = e^{-lambda x}, v = -e^{-lambda x}/lambda$. We have



$$int_{0}^{infty} x^{2} e^{-lambda x} mathop{dx} $$



$$= -frac{x^{2}e^{-lambda x}}{lambda} + int_{0}^{infty} frac{2xe^{-lambda x}}{lambda} mathop{dx} $$






probability probability-theory probability-distributions covariance




probability probability-theory probability-distributions covariance






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 20:52









josephjoseph

500111




500111








  • 2




    $begingroup$
    Best I can do: $$mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY mid X]] = mathbb{E}[Xmathbb{E}[Y mid X]] = mathbb{E}[X^2/2] = dfrac{1}{2}mathbb{E}[X^2]$$
    $endgroup$
    – Clarinetist
    Dec 4 '18 at 20:56










  • $begingroup$
    that will work for me. i can get the variance because i know $X sim text{Exp}(lambda)$. then I can do $mathbb{E}[X^{2}] = (text{Var}(X) + mathbb{E}[X]^{2})$
    $endgroup$
    – joseph
    Dec 4 '18 at 20:58








  • 2




    $begingroup$
    The next step is to simplify even further the computations, noting that $$mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(XE(Ymid X))-E(X)E(E(Ymid X))$$ that is, $$mathrm{Cov}(X,Y)=E(X(tfrac12X))-E(X)E(tfrac12X)=tfrac12mathrm{var}(X)$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:17
















  • 2




    $begingroup$
    Best I can do: $$mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY mid X]] = mathbb{E}[Xmathbb{E}[Y mid X]] = mathbb{E}[X^2/2] = dfrac{1}{2}mathbb{E}[X^2]$$
    $endgroup$
    – Clarinetist
    Dec 4 '18 at 20:56










  • $begingroup$
    that will work for me. i can get the variance because i know $X sim text{Exp}(lambda)$. then I can do $mathbb{E}[X^{2}] = (text{Var}(X) + mathbb{E}[X]^{2})$
    $endgroup$
    – joseph
    Dec 4 '18 at 20:58








  • 2




    $begingroup$
    The next step is to simplify even further the computations, noting that $$mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(XE(Ymid X))-E(X)E(E(Ymid X))$$ that is, $$mathrm{Cov}(X,Y)=E(X(tfrac12X))-E(X)E(tfrac12X)=tfrac12mathrm{var}(X)$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:17










2




2




$begingroup$
Best I can do: $$mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY mid X]] = mathbb{E}[Xmathbb{E}[Y mid X]] = mathbb{E}[X^2/2] = dfrac{1}{2}mathbb{E}[X^2]$$
$endgroup$
– Clarinetist
Dec 4 '18 at 20:56




$begingroup$
Best I can do: $$mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY mid X]] = mathbb{E}[Xmathbb{E}[Y mid X]] = mathbb{E}[X^2/2] = dfrac{1}{2}mathbb{E}[X^2]$$
$endgroup$
– Clarinetist
Dec 4 '18 at 20:56












$begingroup$
that will work for me. i can get the variance because i know $X sim text{Exp}(lambda)$. then I can do $mathbb{E}[X^{2}] = (text{Var}(X) + mathbb{E}[X]^{2})$
$endgroup$
– joseph
Dec 4 '18 at 20:58






$begingroup$
that will work for me. i can get the variance because i know $X sim text{Exp}(lambda)$. then I can do $mathbb{E}[X^{2}] = (text{Var}(X) + mathbb{E}[X]^{2})$
$endgroup$
– joseph
Dec 4 '18 at 20:58






2




2




$begingroup$
The next step is to simplify even further the computations, noting that $$mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(XE(Ymid X))-E(X)E(E(Ymid X))$$ that is, $$mathrm{Cov}(X,Y)=E(X(tfrac12X))-E(X)E(tfrac12X)=tfrac12mathrm{var}(X)$$
$endgroup$
– Did
Dec 4 '18 at 21:17






$begingroup$
The next step is to simplify even further the computations, noting that $$mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(XE(Ymid X))-E(X)E(E(Ymid X))$$ that is, $$mathrm{Cov}(X,Y)=E(X(tfrac12X))-E(X)E(tfrac12X)=tfrac12mathrm{var}(X)$$
$endgroup$
– Did
Dec 4 '18 at 21:17












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