Question

Let X and Y be two independent random variables such that X + Y has the same density as X. What is Y?

Answer 1

A generalised case:

Suppose X and Y are two independent random variables with density functions i.e pdfs fx(x) and fy(y)

Let Z=X+Y

Suppose X takes some specific value integer m. Then Z=z iff Y=z-m.

So the event Z is the union of two disjoint events (X=m) and (Y=z=m)

[note that when two events are independent, P(A∩B)=P(A)*P(B)]

P(Z=z)= ∑P(x=x).P(Y=y) [the range is from -∞ to ∞]

Let this density function be defined as k.

The operation mentioned above is commutative as well as associative [ since the events are independent of each other]

Let Sn=X1+X2+...........+Xn be the sum of n independent random variables of an independent trial process with common density function k defined on the integers.

Such that,

S1=X1

S2=X1+X2

.

.

Sn-1=X1+X2+........Xn-1

Sn=X1+X2+...........+Xn

S1,S2,.........Sn-1,Sn are independent of each other.

i.e. Sn= Sn-1+Xn

thus if we know that the distribution/density of Xn is , we can find the distribution/density of Sn by induction.

i.e. Sn will have the same density function as that of Sn-1 if Xn does not have any value.

HERE,

It is told that X+Y has the same density as that of X, i.e. we can say both (X+Y) and X is defined on the distribution function k. For that to happen, Y also must be degenerated at 0, i.e Y is a degenerated distribution. [otherwise X+Y would have been influenced, and the characteristic of density function would change]

Now what is meant by “degenerated at 0”?

A degenerate distribution is a distribution taking a constant with probability of 1. In other words, a random variable X has a single possible value or constant i.e. value of X is not random in nature or it doesn’t follow any particular characteristic.

For any doubts please ask in the comments.