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CDF, and setting the integral for this

nacho

Active member
Sep 10, 2013
156
Please refer to the attached image.

I can't quite get the bounds for question a) right. it's so confusing. Would it be wise to split the double integral into two parts? I guess thats usually favourable with when dealing with absolute values. But the bounds are still confusing me.

would it be easier to integrate with respect to x first, or w.r.t y?


For x, it would just be 0 to infinity.

for y it would be -infinity to x ? whilst splitting the integral with respect to x? is this correct.
Actually i'll have a crack at it now. some as i make these question threads i tend to find a solution/or at least get closer to it.

Still, help/correction is very much appreciated.
Thanks!

edit: first attempt i did was
$c \int \int xe^{-x}dxdy = 1$ from $0 \ to \ \infty$ and $-\infty \ to \ x$. this was wrong, as it left me with a variable in my answer, so i will try again but int w.r.t y first.

edit 2: This also did not work


edit 3: SOLVED PART A)
i used the bounds -x<y<x and 0<x<infinity
and obtained c= 1/2

edit 4: SOLVED PART B)
I got fy|x(x,y) = $\frac{1}{4x^{3}}$
and fx|y(x,y) = $xe^{-x}$

edit 5: I'm having trouble with part d

E[Y|X=e] = $\int y f_{x|y}(y|x)dy$
 

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chisigma

Well-known member
Feb 13, 2012
1,704
Please refer to the attached image.

I can't quite get the bounds for question a) right. it's so confusing. Would it be wise to split the double integral into two parts? I guess thats usually favourable with when dealing with absolute values. But the bounds are still confusing me.

would it be easier to integrate with respect to x first, or w.r.t y?


For x, it would just be 0 to infinity.

for y it would be -infinity to x ? whilst splitting the integral with respect to x? is this correct.
Actually i'll have a crack at it now. some as i make these question threads i tend to find a solution/or at least get closer to it.

Still, help/correction is very much appreciated.
Thanks!

edit: first attempt i did was
$c \int \int xe^{-x}dxdy = 1$ from $0 \ to \ \infty$ and $-\infty \ to \ x$. this was wrong, as it left me with a variable in my answer, so i will try again but int w.r.t y first.

edit 2: This also did not work


edit 3: SOLVED PART A)
i used the bounds -x<y<x and 0<x<infinity
and obtained c= 1/2

edit 4: SOLVED PART B)
I got fy|x(x,y) = $\frac{1}{4x^{3}}$
and fx|y(x,y) = $xe^{-x}$
What is not full clear is that the p.d.f. is defined as...

$\displaystyle f(x,y)= c\ x\ e^{- x}\ (1)$

... for $\displaystyle x>0,\ |y|< x$ and is 0 in the rest of the x,y plane. In that case the condition for c is...

$\displaystyle c\ \int_{0}^{\infty} \int_{-x}^{+ x} x\ e^{-x}\ d x\ d y = c\ \int_{0}^{\infty} 2\ x^{2}\ e^{- x }\ dx = 4\ c = 1 \implies c= \frac{1}{4}\ (1)$

The remaining part can be attacked with the procedure we already saw...

Kind regards

$\chi$ $\sigma$
 

nacho

Active member
Sep 10, 2013
156
What is not full clear is that the p.d.f. is defined as...

$\displaystyle f(x,y)= c\ x\ e^{- x}\ (1)$

... for $\displaystyle x>0,\ |y|< x$ and is 0 in the rest of the x,y plane. In that case the condition for c is...

$\displaystyle c\ \int_{0}^{\infty} \int_{-x}^{+ x} x\ e^{-x}\ d x\ d y = c\ \int_{0}^{\infty} 2\ x^{2}\ e^{- x }\ dx = 4\ c = 1 \implies c= \frac{1}{4}\ (1)$

The remaining part can be attacked with the procedure we already saw...

Kind regards

$\chi$ $\sigma$
thanks, i had just forgotten to multiply by 2 during a step, leading to me getting 1/2.

can you assist witg part d?i keep getting 0, which i think is wrong
 

chisigma

Well-known member
Feb 13, 2012
1,704
thanks, i had just forgotten to multiply by 2 during a step, leading to me getting 1/2.

can you assist witg part d?i keep getting 0, which i think is wrong
Answering to point d) is performed in three steps...

...first we conpute the marginal distribution of X...

$\displaystyle f_{x} (x) = \frac{1}{4}\ \int_{- x}^{+ x} x\ e^{- x}\ dx = \frac{x^{2}}{2}\ e^{-x}\ (1)$

... second the conditional distribution function of Y...

$\displaystyle f_{y} (y | X=x) = \frac{f(x,y)}{f_{x} (x)} = \frac{1} {2\ x}\ (2)$

... and third...

$\displaystyle E \{Y | X=x \} = \int_{- x}^{ + x} \frac{y}{2\ x}\ d y =0\ (3)$

Kind regards

$\chi$ $\sigma$
 

nacho

Active member
Sep 10, 2013
156
Answering to point d) is performed in three steps...

...first we conpute the marginal distribution of X...

$\displaystyle f_{x} (x) = \frac{1}{4}\ \int_{- x}^{+ x} x\ e^{- x}\ dx = \frac{x^{2}}{2}\ e^{-x}\ (1)$

... second the conditional distribution function of Y...

$\displaystyle f_{y} (y | X=x) = \frac{f(x,y)}{f_{x} (x)} = \frac{1} {2\ x}\ (2)$

... and third...

$\displaystyle E \{Y | X=x \} = \int_{- x}^{ + x} \frac{y}{2\ x}\ d y =0\ (3)$

Kind regards

$\chi$ $\sigma$
ah, i was right then, gues i should be more confident