Expected Value of dependent random Variables

joemama69
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Homework Statement



We are given the following table and need to find the E(XY)

X|Y y = 17 20 23 35 48 p(x)
x = -20 0.02 0.03 0.07 0.02 0.06 0.2
0 0 0.05 0 0.05 0.1 0.2
1 0.05 0.03 0.02 0.07 0.03 0.2
3 0.01 0.02 0.03 0 0.04 0.1
17 0.18 0.04 0.06 0.01 0.01 0.3
p(y) 0.26 0.17 0.18 0.15 0.24 1


Homework Equations





The Attempt at a Solution




So it is clear to me that X & Y are dependent..

E(XY)=SUM_x SUM_y (xyp(x,y)) = -20(17)(.02)+(-20)(20)(.03)... and so on n so forth... is this correct?
 
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joemama69 said:

Homework Statement



We are given the following table and need to find the E(XY)

X|Y y = 17 20 23 35 48 p(x)
x = -20 0.02 0.03 0.07 0.02 0.06 0.2
0 0 0.05 0 0.05 0.1 0.2
1 0.05 0.03 0.02 0.07 0.03 0.2
3 0.01 0.02 0.03 0 0.04 0.1
17 0.18 0.04 0.06 0.01 0.01 0.3
p(y) 0.26 0.17 0.18 0.15 0.24 1


Homework Equations





The Attempt at a Solution




So it is clear to me that X & Y are dependent..

E(XY)=SUM_x SUM_y (xyp(x,y)) = -20(17)(.02)+(-20)(20)(.03)... and so on n so forth... is this correct?

Certainly the formula E(XY) = sum_{x,y} p(x,y) x*y is correct (basically, it is the definition of E(XY)); however, whether or not your numbers are correct is impossible to say, because your table is formatted so badly it is unreadable. You should use tabs, etc., and choose a fixed-width font (such as courier new) for the table. You can do this by selecting the table using the mouse, then go to the "Fonts" menu on the top panel Alternatively, you can use the "tabular" environment in LaTeX/TeX and format your table that way. On-line tutorials will show you how to do it. See, eg., http://www1.maths.leeds.ac.uk/latex/TableHelp1.pdf or
http://www.andy-roberts.net/writing/latex/tables .
 
Last edited:
Whoops, it looks good in the editor but not once posted, didn't realize that, sorry.

That table syntax looks a little confusing... I attached a PDF of all my work. I'm not to confident with my application of the equations. The dependence kind of threw me a bit.
 

Attachments

joemama69 said:

Homework Statement



We are given the following table and need to find the E(XY)
Code: 
X|Y	y = 17	20	23	35	48	p(x)
x = -20	0.02	0.03	0.07	0.02	0.06	0.2
0	0	0.05	0	0.05	0.1	0.2
1	0.05	0.03	0.02	0.07	0.03	0.2
3	0.01	0.02	0.03	0	0.04	0.1
17	0.18	0.04	0.06	0.01	0.01	0.3
p(y)	0.26	0.17	0.18	0.15	0.24	1

Homework Equations





The Attempt at a Solution




So it is clear to me that X & Y are dependent..

E(XY)=SUM_x SUM_y (xyp(x,y)) = -20(17)(.02)+(-20)(20)(.03)... and so on n so forth... is this correct?

Try code tags around your data to preserve spacing like I did above. You probably need to massage it some yet though.
 
LCKurtz said:
Try code tags around your data to preserve spacing like I did above. You probably need to massage it some yet though.

How does one insert code tags?
 
Are you able to use my pdf file w/ my work on it or do I need to retype it in here?
 
Ray Vickson said:
How does one insert code tags?

Just type [ code]
Type your table or text or whatever here. Spaces and line breaks
will be preserved.
[/ code]

I put an extra space in both tags so it wouldn't actually do it. Type the tags without the spaces and the results will look like what you type as in
Code: 
1  2  3
24 25 26
You may still have to tweak it by counting spaces. But see my next post for a better way.
 
Last edited:
Ray Vickson said:
How does one insert code tags?

joemama69 said:
Are you able to use my pdf file w/ my work on it or do I need to retype it in here?

You can actually make a nice table with TeX. Here's one I did an another thread:
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
Q_1Q_2Q_3& J_1=\bar Q_2& K_1= 1& J_2=\bar Q_3& K_2=Q_3& J_3=Q_1& K_3=Q_1\\
\hline
000&1&1&1&0&0&0 \\
\hline
110&&&&&& \\
\hline
&&&&&& \\
\hline
&&&&&& \\
\hline
\end{array}$$Feel free to copy/paste and change it.
 
LCKurtz said:
Just type [ code]
Type your table or text or whatever here. Spaces and line breaks
will be preserved.
[/ code]

I put an extra space in both tags so it wouldn't actually do it. Type the tags without the spaces and the results will look like what you type as in
Code: 
1  2  3
24 25 26
You may still have to tweak it by counting spaces. But see my next post for a better way.

Thank you for the above.
 
  • #10
Code: 
X|Y	y = 17	20	23	35	48	p(x)
x = -20	0.02	0.03	0.07	0.02	0.06	0.2
0	0	0.05	0	0.05	0.1	0.2
1	0.05	0.03	0.02	0.07	0.03	0.2
3	0.01	0.02	0.03	0	0.04	0.1
17	0.18	0.04	0.06	0.01	0.01	0.3
p(y)	0.26	0.17	0.18	0.15	0.24	1

Ok got it, that code trick made it simple thank you... soo back to the problem at hand...

E(X)=∑x p(x)=-20(0.2)+1(0.2)+3(0.1)+17(0.3)=1.6E(Y)=∑y p(y)=17(0.26)+20(0.17)+23(0.18)+35(0.15)+48(0.24)=28.73V(X)=∑(x-E(X))^2 p(x)=

=0.2(-20-1.6)^2+0.2(0-1.6)^2+0.2(1-1.6)^2+0.1(3-1.6)^2

+0.3(17-1.6)^2= 165.24V(Y)=∑(y-E(Y))^2 p(y)=

=0.26(17-28.73)^2+0.17(20-28.73)^2+0.18(23-28.73)^2

+0.15(35-28.73)^2+0.24(48-28.73)^2=149.6571E(XY)=∑_x∑_y xyp(x,y)

=(-20)(17)(0.02)+(-20)(20)(0.03)+(-20)(23)(0.07)+(-20)(35)(0.02)

+(-20)(48)(0.06)+17(0.05)+20(0.03)+35(0.07)+48(0.03)+3(17)(0.01)

+3(20)(0.02)+3(23)(0.03)+3(48)(0.04)+17(17)(0.18)+17(20)(0.04)

+17(23)(0.06)+17(35)(0.01)+17(48)(0.04)=-4.07E(X│Y=23)=∑xp(x│y=23)=∑xp(x,y=23)/p(y=23)

=1/0.18(-20(0.07)+0.02+3(0.03)+17(0.06)=-1.5 V(X│Y=23)=E(X-E(X│Y=23))^2=(1.6+1.5)^2=9.61E(Y│X=-20)=∑yp(y│x=-20)=∑(yp(y,x=-20)/p(x=-20)

= 1/0.2 (17(0.02)+20(0.03)+23(0.07)+35(0.02)+48(0.06))=30.65V(Y│X=-20)=E(Y-E(Y│X=-20))^2=(28.73-30.65)^2=3.6864P(X<3,Y≤23)=0.02+0.03+0.07+0.05+0.05+0.03+0.02=0.27Does it appear that I have used the formulas correctly?
 
Last edited:
  • #11
joemama69 said:
Code: 
X|Y	y = 17	20	23	35	48	p(x)
x = -20	0.02	0.03	0.07	0.02	0.06	0.2
0	0	0.05	0	0.05	0.1	0.2
1	0.05	0.03	0.02	0.07	0.03	0.2
3	0.01	0.02	0.03	0	0.04	0.1
17	0.18	0.04	0.06	0.01	0.01	0.3
p(y)	0.26	0.17	0.18	0.15	0.24	1

Ok got it, that code trick made it simple thank you... soo back to the problem at hand...

E(X)=∑x p(x)=-20(0.2)+1(0.2)+3(0.1)+17(0.3)=1.6


E(Y)=∑y p(y)=17(0.26)+20(0.17)+23(0.18)+35(0.15)+48(0.24)=28.73


V(X)=∑(x-E(X))^2 p(x)=

=0.2(-20-1.6)^2+0.2(0-1.6)^2+0.2(1-1.6)^2+0.1(3-1.6)^2

+0.3(17-1.6)^2= 165.24


V(Y)=∑(y-E(Y))^2 p(y)=

=0.26(17-28.73)^2+0.17(20-28.73)^2+0.18(23-28.73)^2

+0.15(35-28.73)^2+0.24(48-28.73)^2=149.6571


E(XY)=∑_x∑_y xyp(x,y)

=(-20)(17)(0.02)+(-20)(20)(0.03)+(-20)(23)(0.07)+(-20)(35)(0.02)

+(-20)(48)(0.06)+17(0.05)+20(0.03)+35(0.07)+48(0.03)+3(17)(0.01)

+3(20)(0.02)+3(23)(0.03)+3(48)(0.04)+17(17)(0.18)+17(20)(0.04)

+17(23)(0.06)+17(35)(0.01)+17(48)(0.04)=-4.07


E(X│Y=23)=∑xp(x│y=23)=∑xp(x,y=23)/p(y=23)

=1/0.18(-20(0.07)+0.02+3(0.03)+17(0.06)=-1.5


V(X│Y=23)=E(X-E(X│Y=23))^2=(1.6+1.5)^2=9.61


E(Y│X=-20)=∑yp(y│x=-20)=∑(yp(y,x=-20)/p(x=-20)

= 1/0.2 (17(0.02)+20(0.03)+23(0.07)+35(0.02)+48(0.06))=30.65


V(Y│X=-20)=E(Y-E(Y│X=-20))^2=(28.73-30.65)^2=3.6864


P(X<3,Y≤23)=0.02+0.03+0.07+0.05+0.05+0.03+0.02=0.27


Does it appear that I have used the formulas correctly?

The calculation inputs look OK, but I have not done the computations to check the answers.

BTW: a much easier way to calculate variance is ##\text{Var}(X) = E(X^2) - (EX)^2,## so if you compute ##E(X^2)## at the same time as ##EX## you will be almost done. As a useful exercise, you should actually verify this; viz., that
\sum_{x_i} p(x_i) (x_i - EX)^2 = \sum_{x_i} p(x_i) x_i^2 - (EX)^2,
where
EX = \sum_{x_i} p(x_i) x_i.
 
  • #12
Ya we had to verify that in a HW assignment a week or two ago.

I used excel to do all the tedious computations, I was more worried about my set up than the arithmetic.

So for Variance with conditional probabilities I have..

V(X|Y)=E(X-E(X|Y))^2 = (E(X)-E(X|Y))^2 ... is this correct?
 
  • #13
joemama69 said:
Ya we had to verify that in a HW assignment a week or two ago.

I used excel to do all the tedious computations, I was more worried about my set up than the arithmetic.

So for Variance with conditional probabilities I have..

V(X|Y)=E(X-E(X|Y))^2 = (E(X)-E(X|Y))^2 ... is this correct?

No, it is not, and I cannot figure out why you would think it! You get V(X|Y=y) by using the formulas for Var(X), but with p(x) replaced by p_y(x) = Pr(x|Y=y). Note that p_y(x) is a genuine probability distribution over x; it is >= 0, and its sum over all x is 1.
 
  • #14
Ya my professor used that notation but I am not suing it correctly... Guess back to the way I know.V(X│Y=23)=∑(x-E(X|Y=23))^2 p(x|y=23)=(-20+1.5)^2 (0.07/0.18)+(1+1.5)^2 (0.02/0.18)+(3+1.5)^2 (0.03/0.18)+(17+1.5)^2 (0.06/0.18)=1

Am I applying p(x|y) correctly. If I am then I must of done it wrong when I calculated E(X|Y=23)
 

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