Convolution in statistics

• December 27th 2010, 01:56 PM
HHsts
Convolution in statistics
Dear all,

I learned, to derive the common probability distribution function $f_{Z}(i)$
of e.g. two independent distribution functions $f_{X}(i)$X and Y
$f_{X}(i)$, one can use the convolution.

$f_{Z}(i)=\sum_{j=0}^{i}f_{X}(j) f_{Y}(i-j)$

The second way, from my point of view, could be, if e.g. the two distribution have, let's say two states with given probabilities

$f_{X}(i)=[0.6,0.4]$ with 0.6, for instance, as the probability for an electrical current IX1 to occur and 0.4 for an electrical current IX2 to occur.
$f_{Y}(i)=[0.3,0.7]$ with 0.3, for instance, as the probability for an electrical current IY1 to occur and 0.7 for an electrical current IY2 to occur.

The new distribution function of Z is then[tex]
f_{Z}(i=0)=0.6*0.3=0.18
f_{Z}(i=1)=0.6*0.7=0.42
f_{Z}(i=2)=0.4*0.3=0.12
f_{Z}(i=3)=0.4*0.7=0.28[\MATH]
with [tex]f_{Z}(i=1)+f_{Z}(i=2)=0.54[\MATH]

The convolution would give only three stages:

I) $f_{Z}(i=0)=0.6*0.3=0.18$
II) $f_{Z}(i=1)=0.6*0.7+0.4*0.3=0.54$
III) $f_{Z}(i=2)=0.4*0.7=0.28$

The convolution spitts out two results in one as in II). Basically to me that is fine and right, but I do need the fourth stage as I get it without concolution.

Is the second way wrong? From my point of view is it even better as it gives more stages in the result.

So, hopefully one of you guys can help with this.
Which version is correct? From my point of view both.

Thanks

HHsts
• December 27th 2010, 02:14 PM
snowtea
Convolution is a way of deriving probability distribution for Z = X + Y.

I'm not sure exactly what you mean by Z(1), Z(2) since no values were given for X and Y.

Suppose X and Y range over integers 1 through 10 (and X,Y independent)
Asking the probability that Z = 5 for example is the same thing as
P{Z = 5}
= P{X = 1, Y=4} + P{X=3,Y=2} + P{X=3,Y=2} + ...
= P{X = 1}P{Y=4} + P{X=3}P{Y=2} + P{X=3}P{Y=2} + ...

Which is exactly the expression for a convolution.
You can decompose probabilities for P{X=i}P{Y=j}, but adding them gives the same thing as a convolution.
• December 27th 2010, 06:33 PM
mr fantastic
Quote:

Originally Posted by HHsts
Dear all,

I learned, to derive the common probability distribution function $f_{Z}(i)$
of e.g. two independent distribution functions $f_{X}(i)$X and Y
$f_{X}(i)$, one can use the convolution.

$f_{Z}(i)=\sum_{j=0}^{i}f_{X}(j) f_{Y}(i-j)$

The second way, from my point of view, could be, if e.g. the two distribution have, let's say two states with given probabilities

X=[0.6 0.4];
Y=[0.3 0.7];

Z(1)=0.6*0.3=0.18
Z(2)=0.6*0.7=0.42
Z(3)=0.4*0.3=0.12
Z(4)=0.4*0.7=0.28
with Z(2)+Z(3)=0.54

The convolution would give:

0.1800 0.5400 0.2800

Is the second way wrong? From my point of view is it even better as it gives more stages in the result.

So, hopefully one of you guys can help with this.
Which version is correct? From my point of view both.

Thanks

HHsts

Please post the entire original question you are working on, not just your summary of it.
• December 28th 2010, 12:54 AM
HHsts
Hello snowtea,

Thanks for your reply. I have detailed my question (see original post). Hope this does clarify my concern.

HHsts
• December 28th 2010, 03:26 AM
snowtea
$f_Z(1) = f_X(1)f_Y(0)+f_X(0)f_Y(1)$

This is the correct probability. For Z=1, we can have either X=1 Y=0 or X=0 Y=1.
This is the expression given by the convolution. Dividing it into "multiple stages" does not help when you just want probabilities for a particular value of Z.
• December 29th 2010, 03:01 AM
HHsts
Snowtea,

I see that Z=1 can happen if X=1 and Y=0 or Y=1 and X=0.
according to
X|Y|Z
0|0|0
1|0|1
0|1|1
1|1|1

if you mean this and apply binomial distr.

I still have one thing on my mind.

I have four current, as above

System X) Can take IX1=3A with 0.6 probability to occur and IX2=5A with 0.4 prob. to occur
System Y) can take IY1=1A with 0.3 to occur and IY2=2A with 0.7 prob. to occur

Now, I can get four current levels.

1) IX1+IY1=3A+1A=4A
2) IX1+IY2=3A+2A=5A
3) IX2+IY1=5A+1A=6A
4) IX2+IY2=5A+2A=7A

When I would like to find the prob. for each level to occur I require four results

1) IX1+IY1=3A+1A=4A=0.6*0.3
2) IX1+IY2=3A+2A=5A=0.6*0.7
3) IX2+IY1=5A+1A=6A=0.4*0.3
4) IX2+IY2=5A+2A=7A=0.4*0.7

The convolution would give me 1) then 2+3 and then 4.

But I would like to find all four seperate prob.s
So what do?

Thanks again for a short feedback.

HHsts