Can Two Different Random Variables Have the Same Distribution?

[wuid]

Hello ,

frustrated with my lecturer assignments , i need your help with this :

if X,Y and are two different random variable , is it possible that X,X+Y have equally distributed.

if it can be give an example , if not prove it.

thanks,

 

[Ray Vickson]

Hello ,

frustrated with my lecturer assignments , i need your help with this :

if X,Y and are two different random variable , is it possible that X,X+Y have equally distributed.

if it can be give an example , if not prove it.

thanks,

Show your work---those are the Forum rules.

RGV

 

[wuid]

sorry but i don't have any ,
i don't remember that we learned who to prove equality in distribution.
all i thought of is that if i want to prove it right i have to prove that :
P(X$\leq$t)=P(X+Y$\leq$t) for all t,

maybe there other conditions that lead to equality in distribution, but i don't know them.

generally i think it is possible , so i am trying to think on distribution that will satisfy the answer

 

[Ray Vickson]

sorry but i don't have any ,
i don't remember that we learned who to prove equality in distribution.
all i thought of is that if i want to prove it right i have to prove that :
P(X$\leq$t)=P(X+Y$\leq$t) for all t,

maybe there other conditions that lead to equality in distribution, but i don't know them.

generally i think it is possible , so i am trying to think on distribution that will satisfy the answer

Equality in distribution is equivalent to equality of the characteristic functions,or Laplace transforms, or moment-generating functions, etc.

RGV

 

[wuid]

those subjects you mentioned aren't in the topics of the course , haven't learned them...

maybe a simple case that answer this problem ?

 

[jbunniii]

Are there any constraints at all on X and Y?

For example, if X is equal to some constant with probability 1, can you find a solution in that case?

 

[wuid]

jbunniii , if i understood you ,
if x is a constant with probability 1 and y is also the same constant with probability 1 then
x,x+y has the same distribution.

is it right ?

but the only constraint about x and y that they will be different

 

[jbunniii]

jbunniii , if i understood you ,
if x is a constant with probability 1 and y is also the same constant with probability 1 then
x,x+y has the same distribution.

is it right ?

Well, it depends on which constant you choose. If x and y are constants, and you need x = x + y, then what does this imply?

 

[wuid]

going back to the start , if X,Y are different random variable they can't be constants.
no ? i can't see where is the randomness of X or Y if the are constants.

 

[jbunniii]

Certainly a random variable can be constant. Simply assign it a constant value with probability 1. Yes, it's a somewhat trivial example, but that's why I asked if there are any constraints on what kind of distribution can be assumed.

To find out if there is a less trivial example, consider this: if X and X + Y have the same distribution, then they must have the same moments. Try using this fact for the first and second moments, and see what you can conclude.

 

[wuid]

if i take a fair coin where , X=1 for heads , X=0 for tail so P(X=1)=P(X=0)=0.5 ,
and i will define Y=1-X so P(Y=1)=P(Y=0)=0.5 ,
can i say that P(X=0)=P(X=1)=P(X+Y=1)=0.5 ?
this means X,X+Y have equal distribution and X,Y are not constants ?

 

[jbunniii]

if i take a fair coin where , X=1 for heads , X=0 for tail so P(X=1)=P(X=0)=0.5 ,
and i will define Y=1-X so P(Y=1)=P(Y=0)=0.5 ,
can i say that P(X=0)=P(X=1)=P(X+Y=1)=0.5 ?
this means X,X+Y have equal distribution and X,Y are not constants ?

I don't think this example works. If Y = 1 - X, then X + Y cannot equal 0, whereas X can, with probability 0.5. Therefore X and X + Y do not have the same probability distribution.

 

[wuid]

o.k got it , with slight modification - Z=1-X , Y=1-2*X then Z=X+Y, where X is Bernoulli
=> X,Z equal in distribution.

 

[jbunniii]

o.k got it , with slight modification - Z=1-X , Y=1-2*X then Z=X+Y, where X is Bernoulli
=> X,Z equal in distribution.

Looks good to me.