Probability problem -- The joint PDF of X and Y is uniform on this rectangle....

[docnet]

Homework Statement

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Relevant Equations

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we haven't learned howw to do parts d-f yest. could you please give me a hand?

(a)
$$E(X)=E(Y)=\int_0^3\int_0^3\frac{1}{9}dxdy=\frac{3}{2}$$
(b)its typoe suppoesed to be W=Y-2
$$E(Z)=E(X-2)=E(X)-E(2)=-\frac{1}{2}$$
$$E(W)=E(Y-2)=E(Y)-E(2)=-\frac{1}{2}$$
(c) i guess joint pdf from $E(Z)$ and $E(W)$
$$ f_{ZW}=\frac{1}{9}\quad\text{for}\quad -2\leq a\leq 1, -2\leq b\leq 1$$
(d) is there a rule or a theorem ew can use her? this odens't mke much sense to me.
(e) if we have the joint pdf, could we integrate along w to get $f_T$?
(f) $E(T)=\int f_T ds$ or $\int (x+y-4) f_T ds$ because I am not suress. pls help

 

[Orodruin]

(d) What is the definition of the pdf in two variables? How can you use this to determine how the pdf changes when you change variables?

 

[FactChecker]

a) The size of the rectangle is 2 by 2, not 3 by 3.
b) The question asks for the pdf, not the expected values.
c) Correct this for the error in the size of the rectangle. Then compare it to the original pdf.
d) I don't think you would call it a "rule or theorem". As an example, what values of W and Z would give W=0, T=0? Then what is the pdf value of $f_{W,Z}$ at that point? Is that the same as the value of $f_{W,T}(0,0)$?
Use this example for general pdf values of W and T.
(WARNING: This simple approach only works in this case because the probabilities of W and Z are not being spread over a different range. For a problem like T=Z+2W, the 2 multiplier would reduce the pdf. You would need to account for that.)
e) Yes.
d) You should probably get the prior parts so that you can fill in actual values and limits in the integral. I am not familiar with LOTUS. I'll defer to others on this.

 

[docnet]

(d) What is the definition of the pdf in two variables? How can you use this to determine how the pdf changes when you change variables?

in my professor's notes the pdf in two variables are defined as $$f_{XY}=\frac{\partial ^2F_{XY}}{\partial_x\partial_y}$$ I'm not sure what to do with this information because the CDF in two variables seems as difficult to come up with.

a) The size of the rectangle is 2 by 2, not 3 by 3.
b) The question asks for the pdf, not the expected values.
c) Correct this for the error in the size of the rectangle. Then compare it to the original pdf.
d) I don't think you would call it a "rule or theorem". As an example, what values of W and Z would give W=0, T=0? Then what is the pdf value of $f_{W,Z}$ at that point? Is that the same as the value of $f_{W,T}(0,0)$?
Use this example for general pdf values of W and T.
(WARNING: This simple approach only works in this case because the probabilities of W and Z are not being spread over a different range. For a problem like T=Z+2W, the 2 multiplier would reduce the pdf. You would need to account for that.)
e) Yes.
d) You should probably get the prior parts so that you can fill in actual values and limits in the integral. I am not familiar with LOTUS. I'll defer to others on this.

(a) the rectangle is 2 by two, so area 4. so $$f_{XY}=\frac{1}{4}$$ leading to $$E(X)=E(Y)=\frac{1}{2}$$

(b) let the new variable be $c=a-2$ and $d=b-2$
$$ f_{ZW}=\frac{1}{4} $$ in the square $-1\leq c \leq 1$ and $-1\leq d \leq 1$

(c) $f_{ZW}$ is like $f_{XY}$ translated by $(-2,2)$

(d) let $e=c+d$ then $-2\leq e \leq 2$

shakey :

i try finding the pdf $f_T$. the marginal pdfs $f_W$ and $f_Z$ are constant, so their sum is constant (pretty sure this is wrong way to find the pdf of T and it isn't constant) leading to$f_{T}=\frac{1}{4}$ on the interval $-2\leq e \leq 2$.

leading to another constant joint pdf $$f_{TW}=\frac{1}{8}$$

(e) $f_{T}=\frac{1}{4}$ on the interval $-2\leq e \leq 2$ by earlier erranous result

(d) T is summetryc about 0, so its expected value (mean) is 0. using lotus its like $$\int_{-2}^2f_{T}de=\int_{1}^3\int_{1}^3((x-2)+(y-2))\frac{1}{4}dxdy= 0$$ something is wrounge

 

[FactChecker]

For d) you are jumping to the marginal probability of T unnecessarily. When you try to find something like $f_{W,T}(0.1,0)$ you know that $W=0.1$, which had a pdf, $f_W(0.1) = 1/2$. Given that, you also know that $Z=-0.1$, which had a pdf $f_Z(-0.1) = 1/2$. Since $W$ and $Z$ are independent, you can multiply their pdfs together to get the joint pdf, $f_{W,T}(0.1,0)=f_W(0.1)*f_Z(-0.1)=1/2*1/2=1/4$

 

[docnet]

For d) you are jumping to the marginal probability of T unnecessarily. When you try to find something like $f_{W,T}(0.1,0)$ you know that $W=0.1$, which had a pdf, $f_W(0.1) = 1/2$. Given that, you also know that $Z=-0.1$, which had a pdf $f_Z(-0.1) = 1/2$. Since $W$ and $Z$ are independent, you can multiply their pdfs together to get the joint pdf, $f_{W,T}(0.1,0)=f_W(0.1)*f_Z(-0.1)=1/2*1/2=1/4$

omg this clears up so much confusion. thank you!

so the marginal pdf of T is $\frac{1}{2}$ on the interval $[-1,1]$ ??

 

[docnet]

Since $W$ and $Z$ are independent, you can multiply their pdfs together to get the joint pdf, ##f_{W,T}(0.1,0)=f_W(0.1)*f_Z(-0.1)=1/2*1/2=1/4#

would you please explain how $f_{Z+W}(0) =f_Z(-1)$? i understand that Z=-1 and W=1 makes Z+W=T=0, but i don't understand how $f_{Z+W}$ can be equal to $f_Z$

 

[docnet]

oh sorry i think i got it. because $W=.1$ and $T=0$, $Z$ has to be $-.1$. so $f_{Z+.1=0}=f_{Z=-.1}$ except the notation :) is confuseing

 

[FactChecker]

would you please explain how $f_{Z+W}(0) =f_Z(-1)$? i understand that Z=-1 and W=1 makes Z+W=T=0, but i don't understand how $f_{Z+W}$ can be equal to $f_Z$

I don't think I ever said that. T and W are not independent, but Z and W are. We know the pdfs of both Z and W from their original definitions based on X and Y, respectively.

 

[FactChecker]

would you please explain how $f_{Z+W}(0) =f_Z(-1)$? i understand that Z=-1 and W=1 makes Z+W=T=0, but i don't understand how $f_{Z+W}$ can be equal to $f_Z$

Just to be more clear, it is the conditional pdf, $f_{Z+W|W=0.1}(0)$ that is equal to $f_Z(-0.1)$. Once you know that $W=0.1$ the probability of $Z+W$ is directly related to the probability of $Z$.

 

[Orodruin]

Since (d) has been essentially resolved, I would like to offer an alternative approach.

The pdf $f_{XY}(x,y)$ is the function such that $f_{XY}(x,y) dx\, dy$ is the probability that $X$ is between $x$ and $x+dx$ and $Y$ is between $y$ and $y + dy$. With the corresponding definition for a transformed set of stochastic variables $Z$ and $W$, you would therefore find that
$$
f_{XY}(x,y) dx\, dy = f_{ZW}(z,w) dz\, dw.
$$
This just corresponds to a regular change of variables in a two-dimensional integral and the two pdfs are therefore related by the Jacobian determinant of the change of variables.

In this case, changing from $(z,w)$ to $(t,w) = (z+w,w)$, the Jacobian determinant is one and the pdfs are therefore equal (on corresponding values of $(t,w)$ and $(z,w)$).

Additional note: The problem as written specifies $Y = W-2$, not $W = Y - 2$. It is likely a typo, but as written has a different solution.

 

[docnet]

using @Orodruin's method of the Jacobian determinant, $f_{T,W}=\frac{1}{4}$

the corresponding region in the $t-w$ plane is the parallelogram with vertices at $(0,1), (2,1), (0,-1),$ and $(-2,-1)$.

$$f_T=\begin{cases}\int^{t+1}_{-1}\frac{1}{4}dw=\frac{t+2}{4}, -2\leq t\leq 0\\
\int_{t-1}^1\frac{1}{4}dw=\frac{2-t}{4}, 0\leq t\leq 2 \\ 0 \quad otherwise\end{cases}$$

the $f_T$ is constant on the parallalogram, and the parrallogram has equal areas on either side of $t=0$ so the the expected (mean value weighed according to probability) is zero. $$\int_{-2}^2 \frac{t}{4}dt=\int_{-1}^1\int_{-1}^1\frac{(w+z)}{4}dwdz=0$$