Mixing
Determine the joint density function $f(x,y)$ for $0<y0$
$begingroup$
The continuous random variables X and Y have the joint distribution function
$$F(x,y)=begin{cases}1-e^{-x} & y>1 ,:x>0\y(1-e^{-x}) & 0<y<1,:x > 0\0 & elseend{cases}$$
Determine the joint density function $f(x,y)$ for $0<y<1$ and $x>0$
Usually to go from the distribution function to the density function I differentiate the distribution function, but in this case I have the joint so it depends on $x$ and $y$, how can I calculate the joint density function? Also because the interval $x>0$ is present in $2$ cases, how can I deal with?
probability statistics
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
$endgroup$
add a comment |
$begingroup$
The continuous random variables X and Y have the joint distribution function
$$F(x,y)=begin{cases}1-e^{-x} & y>1 ,:x>0\y(1-e^{-x}) & 0<y<1,:x > 0\0 & elseend{cases}$$
Determine the joint density function $f(x,y)$ for $0<y<1$ and $x>0$
Usually to go from the distribution function to the density function I differentiate the distribution function, but in this case I have the joint so it depends on $x$ and $y$, how can I calculate the joint density function? Also because the interval $x>0$ is present in $2$ cases, how can I deal with?
probability statistics
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
$endgroup$
add a comment |
$begingroup$
The continuous random variables X and Y have the joint distribution function
$$F(x,y)=begin{cases}1-e^{-x} & y>1 ,:x>0\y(1-e^{-x}) & 0<y<1,:x > 0\0 & elseend{cases}$$
Determine the joint density function $f(x,y)$ for $0<y<1$ and $x>0$
Usually to go from the distribution function to the density function I differentiate the distribution function, but in this case I have the joint so it depends on $x$ and $y$, how can I calculate the joint density function? Also because the interval $x>0$ is present in $2$ cases, how can I deal with?
probability statistics
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
$endgroup$
The continuous random variables X and Y have the joint distribution function
$$F(x,y)=begin{cases}1-e^{-x} & y>1 ,:x>0\y(1-e^{-x}) & 0<y<1,:x > 0\0 & elseend{cases}$$
Determine the joint density function $f(x,y)$ for $0<y<1$ and $x>0$
Usually to go from the distribution function to the density function I differentiate the distribution function, but in this case I have the joint so it depends on $x$ and $y$, how can I calculate the joint density function? Also because the interval $x>0$ is present in $2$ cases, how can I deal with?
probability statistics
probability statistics
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
asked Jan 26 at 17:15
Luke MarciLuke Marci
856
856
add a comment |
add a comment |
1 Answer
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$begingroup$
You simply differentiate with respect to both variables, i.e. $$frac{partial F(x,y)}{partial x partial y}=begin{cases}0 & y>1 ,:x>0\e^{-x} & 0<y<1,:x > 0\0 & elseend{cases}$$
The answer might seem counter-intuitive but inspect the joint CDF. When $y>1$, $F(x,y)=P(Xleq x, Y leq y)$ doesn't change with respect to $y$, i.e. there is no volume for $y$, when $y>1$.
answered Jan 26 at 18:34
gunesgunes
3747
$endgroup$
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You simply differentiate with respect to both variables, i.e. $$frac{partial F(x,y)}{partial x partial y}=begin{cases}0 & y>1 ,:x>0\e^{-x} & 0<y<1,:x > 0\0 & elseend{cases}$$
The answer might seem counter-intuitive but inspect the joint CDF. When $y>1$, $F(x,y)=P(Xleq x, Y leq y)$ doesn't change with respect to $y$, i.e. there is no volume for $y$, when $y>1$.
answered Jan 26 at 18:34
gunesgunes
3747
$endgroup$
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
add a comment |
$begingroup$
You simply differentiate with respect to both variables, i.e. $$frac{partial F(x,y)}{partial x partial y}=begin{cases}0 & y>1 ,:x>0\e^{-x} & 0<y<1,:x > 0\0 & elseend{cases}$$
The answer might seem counter-intuitive but inspect the joint CDF. When $y>1$, $F(x,y)=P(Xleq x, Y leq y)$ doesn't change with respect to $y$, i.e. there is no volume for $y$, when $y>1$.
answered Jan 26 at 18:34
gunesgunes
3747
$endgroup$
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
add a comment |
$begingroup$
You simply differentiate with respect to both variables, i.e. $$frac{partial F(x,y)}{partial x partial y}=begin{cases}0 & y>1 ,:x>0\e^{-x} & 0<y<1,:x > 0\0 & elseend{cases}$$
The answer might seem counter-intuitive but inspect the joint CDF. When $y>1$, $F(x,y)=P(Xleq x, Y leq y)$ doesn't change with respect to $y$, i.e. there is no volume for $y$, when $y>1$.
answered Jan 26 at 18:34
gunesgunes
3747
$endgroup$
You simply differentiate with respect to both variables, i.e. $$frac{partial F(x,y)}{partial x partial y}=begin{cases}0 & y>1 ,:x>0\e^{-x} & 0<y<1,:x > 0\0 & elseend{cases}$$
The answer might seem counter-intuitive but inspect the joint CDF. When $y>1$, $F(x,y)=P(Xleq x, Y leq y)$ doesn't change with respect to $y$, i.e. there is no volume for $y$, when $y>1$.
answered Jan 26 at 18:34
gunesgunes
3747
answered Jan 26 at 18:34
gunesgunes
3747
answered Jan 26 at 18:34
gunesgunes
3747
answered Jan 26 at 18:34
gunesgunes
3747
3747
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
add a comment |
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
$begingroup$
You are right, thank you a lot for the answer! :)
$endgroup$
– Luke Marci
Jan 26 at 18:44
add a comment |
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