Mixing
Let $X$ and $Y$ be two continuous random variables with joint probability density function
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Let $X$ and $Y$ be two continuous random variables with joint probability density function $f_{XY}(x,y)=Kcdot(xcdot y+1)$ for $0≤x≤1$ and $0≤y≤x$, and $f_{XY}(x,y)=0$ otherwise.
a) Prove that, since $f_{XY}(x,y)$ is a joint probability density function, $K$ equals $frac{8}{5}$.
b) Show that the marginal density function of $X$ is $f_X(x)=frac{8}{5}cdot(frac{X^3}{2}+x)$ if $0≤x≤1$ and $f_X(x)=0$ otherwise.
c) Use (b) to compute the distribution function of $X$.
d) Compute $E[XY]$.
Solve:
a) $F_{X,Y}(x,y)=int_{0}^{1} int_{0}^{x}Kcdot(xcdot y+1)dydx=1$ so $K=frac{8}{5}$ What I compute is $F_{X,Y}(x,y)$ or should I call it in onother way (maybe just $F(x,y)$)?
b) $f_{X}(x)=int_{0}^{x}frac{8}{5}cdot(xcdot y +1)dydx=frac{8}{5}cdot(frac{x^3}{2} +x)$, same as above, what I compute is $f_{X}(x)$ or should I call it in onother way?
c) I don't know how to compute it, I think is the derivative of what I calculated in b.
d) I think that it is $E[XY]=int_{0}^{x} xcdot y cdot f(x,y) dy$ or $E[XY]=int_{0}^{x} xcdot y cdot f_x(x,y) dy$.
Is not clear for me what is the joint probability density function, marginal density function, distribution function in symbols, like the difference between $f$, $f_x$ and $F_X$ can someone please explain to me? And also can someone help me with this exercise. Thanks!
probability
asked Jan 20 at 19:00
FTACFTAC
2649
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be two continuous random variables with joint probability density function $f_{XY}(x,y)=Kcdot(xcdot y+1)$ for $0≤x≤1$ and $0≤y≤x$, and $f_{XY}(x,y)=0$ otherwise.
a) Prove that, since $f_{XY}(x,y)$ is a joint probability density function, $K$ equals $frac{8}{5}$.
b) Show that the marginal density function of $X$ is $f_X(x)=frac{8}{5}cdot(frac{X^3}{2}+x)$ if $0≤x≤1$ and $f_X(x)=0$ otherwise.
c) Use (b) to compute the distribution function of $X$.
d) Compute $E[XY]$.
Solve:
a) $F_{X,Y}(x,y)=int_{0}^{1} int_{0}^{x}Kcdot(xcdot y+1)dydx=1$ so $K=frac{8}{5}$ What I compute is $F_{X,Y}(x,y)$ or should I call it in onother way (maybe just $F(x,y)$)?
b) $f_{X}(x)=int_{0}^{x}frac{8}{5}cdot(xcdot y +1)dydx=frac{8}{5}cdot(frac{x^3}{2} +x)$, same as above, what I compute is $f_{X}(x)$ or should I call it in onother way?
c) I don't know how to compute it, I think is the derivative of what I calculated in b.
d) I think that it is $E[XY]=int_{0}^{x} xcdot y cdot f(x,y) dy$ or $E[XY]=int_{0}^{x} xcdot y cdot f_x(x,y) dy$.
Is not clear for me what is the joint probability density function, marginal density function, distribution function in symbols, like the difference between $f$, $f_x$ and $F_X$ can someone please explain to me? And also can someone help me with this exercise. Thanks!
probability
asked Jan 20 at 19:00
FTACFTAC
2649
$endgroup$
2
$begingroup$
For c), integrate the result in b) from $0$ to $x$ which will give you the distribution function (cdf - $F_X(x)$). For d), integrate the first integral $(xyf_{X,Y}(x,y)dxdy)$ on the entire range $(0≤y≤x, 0≤x≤1)$.
$endgroup$
– Sauhard Sharma
Jan 20 at 19:16
$begingroup$
thanks a lot man!
$endgroup$
– FTAC
Jan 20 at 20:24
add a comment |
$begingroup$
Let $X$ and $Y$ be two continuous random variables with joint probability density function $f_{XY}(x,y)=Kcdot(xcdot y+1)$ for $0≤x≤1$ and $0≤y≤x$, and $f_{XY}(x,y)=0$ otherwise.
a) Prove that, since $f_{XY}(x,y)$ is a joint probability density function, $K$ equals $frac{8}{5}$.
b) Show that the marginal density function of $X$ is $f_X(x)=frac{8}{5}cdot(frac{X^3}{2}+x)$ if $0≤x≤1$ and $f_X(x)=0$ otherwise.
c) Use (b) to compute the distribution function of $X$.
d) Compute $E[XY]$.
Solve:
a) $F_{X,Y}(x,y)=int_{0}^{1} int_{0}^{x}Kcdot(xcdot y+1)dydx=1$ so $K=frac{8}{5}$ What I compute is $F_{X,Y}(x,y)$ or should I call it in onother way (maybe just $F(x,y)$)?
b) $f_{X}(x)=int_{0}^{x}frac{8}{5}cdot(xcdot y +1)dydx=frac{8}{5}cdot(frac{x^3}{2} +x)$, same as above, what I compute is $f_{X}(x)$ or should I call it in onother way?
c) I don't know how to compute it, I think is the derivative of what I calculated in b.
d) I think that it is $E[XY]=int_{0}^{x} xcdot y cdot f(x,y) dy$ or $E[XY]=int_{0}^{x} xcdot y cdot f_x(x,y) dy$.
Is not clear for me what is the joint probability density function, marginal density function, distribution function in symbols, like the difference between $f$, $f_x$ and $F_X$ can someone please explain to me? And also can someone help me with this exercise. Thanks!
probability
asked Jan 20 at 19:00
FTACFTAC
2649
$endgroup$
Let $X$ and $Y$ be two continuous random variables with joint probability density function $f_{XY}(x,y)=Kcdot(xcdot y+1)$ for $0≤x≤1$ and $0≤y≤x$, and $f_{XY}(x,y)=0$ otherwise.
a) Prove that, since $f_{XY}(x,y)$ is a joint probability density function, $K$ equals $frac{8}{5}$.
b) Show that the marginal density function of $X$ is $f_X(x)=frac{8}{5}cdot(frac{X^3}{2}+x)$ if $0≤x≤1$ and $f_X(x)=0$ otherwise.
c) Use (b) to compute the distribution function of $X$.
d) Compute $E[XY]$.
Solve:
a) $F_{X,Y}(x,y)=int_{0}^{1} int_{0}^{x}Kcdot(xcdot y+1)dydx=1$ so $K=frac{8}{5}$ What I compute is $F_{X,Y}(x,y)$ or should I call it in onother way (maybe just $F(x,y)$)?
b) $f_{X}(x)=int_{0}^{x}frac{8}{5}cdot(xcdot y +1)dydx=frac{8}{5}cdot(frac{x^3}{2} +x)$, same as above, what I compute is $f_{X}(x)$ or should I call it in onother way?
c) I don't know how to compute it, I think is the derivative of what I calculated in b.
d) I think that it is $E[XY]=int_{0}^{x} xcdot y cdot f(x,y) dy$ or $E[XY]=int_{0}^{x} xcdot y cdot f_x(x,y) dy$.
Is not clear for me what is the joint probability density function, marginal density function, distribution function in symbols, like the difference between $f$, $f_x$ and $F_X$ can someone please explain to me? And also can someone help me with this exercise. Thanks!
probability
probability
asked Jan 20 at 19:00
FTACFTAC
2649
asked Jan 20 at 19:00
FTACFTAC
2649
asked Jan 20 at 19:00
FTACFTAC
2649
asked Jan 20 at 19:00
FTACFTAC
2649
asked Jan 20 at 19:00
FTACFTAC
2649
2649
2
$begingroup$
For c), integrate the result in b) from $0$ to $x$ which will give you the distribution function (cdf - $F_X(x)$). For d), integrate the first integral $(xyf_{X,Y}(x,y)dxdy)$ on the entire range $(0≤y≤x, 0≤x≤1)$.
$endgroup$
– Sauhard Sharma
Jan 20 at 19:16
$begingroup$
thanks a lot man!
$endgroup$
– FTAC
Jan 20 at 20:24
add a comment |
2
$begingroup$
For c), integrate the result in b) from $0$ to $x$ which will give you the distribution function (cdf - $F_X(x)$). For d), integrate the first integral $(xyf_{X,Y}(x,y)dxdy)$ on the entire range $(0≤y≤x, 0≤x≤1)$.
$endgroup$
– Sauhard Sharma
Jan 20 at 19:16
$begingroup$
thanks a lot man!
$endgroup$
– FTAC
Jan 20 at 20:24
2
2
$begingroup$
For c), integrate the result in b) from $0$ to $x$ which will give you the distribution function (cdf - $F_X(x)$). For d), integrate the first integral $(xyf_{X,Y}(x,y)dxdy)$ on the entire range $(0≤y≤x, 0≤x≤1)$.
$endgroup$
– Sauhard Sharma
Jan 20 at 19:16
$begingroup$
For c), integrate the result in b) from $0$ to $x$ which will give you the distribution function (cdf - $F_X(x)$). For d), integrate the first integral $(xyf_{X,Y}(x,y)dxdy)$ on the entire range $(0≤y≤x, 0≤x≤1)$.
$endgroup$
– Sauhard Sharma
Jan 20 at 19:16
$begingroup$
thanks a lot man!
$endgroup$
– FTAC
Jan 20 at 20:24
$begingroup$
thanks a lot man!
$endgroup$
– FTAC
Jan 20 at 20:24
add a comment |
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$begingroup$
For c), integrate the result in b) from $0$ to $x$ which will give you the distribution function (cdf - $F_X(x)$). For d), integrate the first integral $(xyf_{X,Y}(x,y)dxdy)$ on the entire range $(0≤y≤x, 0≤x≤1)$.
$endgroup$
– Sauhard Sharma
Jan 20 at 19:16
$begingroup$
thanks a lot man!
$endgroup$
– FTAC
Jan 20 at 20:24