is said to be a compound random variable, with the distribution of N called the compounding distribution. In this subsection we will first derive an identity involving such random variables. We will then specialize to where the Xi are positive integer valued random variables, prove a corollary of the identity, and then use this corollary to develop a recursive formula for the probability mass function of SN, for a variety of common compounding distributions.

To begin, let M be a random variable that is independent of the sequence X1,X2,…, and which is such that

P{M=n}=nP{N=n}E[N],n=1,2,…

Proposition 3.5

The Compound Random Variable Identity

For any function h

E[SNh(SN)]=E[N]E[X1h(SM)]

Proof

E[SNh(SN)]=E∑i=1NXih(SN)=∑n=0∞E∑i=1NXih(SN)∣N=nP{N=n}(by conditioning onN)=∑n=0∞E∑i=1nXih(Sn)∣N=nP{N=n}=∑n=0∞E∑i=1nXih(Sn)P{N=n}(by independence ofNandX1,…,Xn)=∑n=0∞∑i=1nEXih(Sn)P{N=n}

Now, because X1,…,Xn are independent and identically distributed, and h(Sn)=h(X1+⋯+Xn) is a symmetric function of X1,…,Xn, it follows that the distribution of Xih(Sn) is the same for all i=1,…,n. Therefore, continuing the preceding string of equalities yields

E[SNh(SN)]=∑n=0∞nE[X1h(Sn)]P{N=n}=E[N]∑n=0∞E[X1h(Sn)]P{M=n}(definition ofM)=E[N]∑n=0∞E[X1h(Sn)∣M=n]P{M=n}(independence ofMandX1,…,Xn)=E[N]∑n=0∞E[X1h(SM)∣M=n]P{M=n}=E[N]E[X1h(SM)]

which proves the proposition. ■

Suppose now that the Xi are positive integer valued random variables, and let

αj=P{X1=j},j>0

The successive values of P{SN=k} can often be obtained from the following corollary to Proposition 3.5.

Corollary 3.6

P{SN=0}=P{N=0}P{SN=k}=1kE[N]∑j=1kjαjP{SM-1=k-j},k>0

Proof

For k fixed, let

h(x)=1,ifx=k0,ifx≠k

and note that SNh(SN) is either equal to k if SN=k or is equal to 0 otherwise. Therefore,

E[SNh(SN)]=kP{SN=k}

and the compound identity yields

kP{SN=k}=E[N]E[X1h(SM)]=E[N]∑j=1∞E[X1h(SM))∣X1=j]αj=E[N]∑j=1∞jE[h(SM)∣X1=j]αj=E[N]∑j=1∞jP{SM=k∣X1=j}αj (3.37)

(3.37)

Now,

P{SM=k∣X1=j}=P∑i=1MXi=kX1=j=Pj+∑i=2MXi=kX1=j=Pj+∑i=2MXi=k=Pj+∑i=1M-1Xi=k=P{SM-1=k-j}

The next to last equality followed because X2,…,XM and X1,…,XM-1 have the same joint distribution; namely that of M-1 independent random variables that all have the distribution of X1, where M-1 is independent of these random variables. Thus the proof follows from Equation (3.37). ■

When the distributions of M-1 and N are related, the preceding corollary can be a useful recursion for computing the probability mass function of SN, as is illustrated in the following subsections.

3.7.1 Poisson Compounding Distribution

If N is the Poisson distribution with mean λ, then

P{M-1=n}=P{M=n+1}=(n+1)P{N=n+1}E[N]=1λ(n+1)e-λλn+1(n+1)!=e-λλnn!

Consequently, M-1 is also Poisson with mean λ. Thus, with

Pn=P{SN=n}

the recursion given by Corollary 3.6 can be written

P0=e-λPk=λk∑j=1kjαjPk-j,k>0

Remark

When the Xi are identically 1, the preceding recursion reduces to the well-known identity for a Poisson random variable having mean λ:

P{N=0}=e-λP{N=n}=λnP{N=n-1},n⩾1

Example 3.34

Let S be a compound Poisson random variable with λ=4 and

P{Xi=i}=1/4,i=1,2,3,4

Let us use the recursion given by Corollary 3.6 to determine P{S=5}. It gives

P0=e-λ=e-4P1=λα1P0=e-4P2=λ2(α1P1+2α2P0)=32e-4P3=λ3(α1P2+2α2P1+3α3P0)=136e-4P4=λ4(α1P3+2α2P2+3α3P1+4α4P0)=7324e-4P5=λ5(α1P4+2α2P3+3α3P2+4α4P1+5α5P0)=381120e-4■

3.7.2 Binomial Compounding Distribution

Suppose that N is a binomial random variable with parameters r and p. Then,

P{M-1=n}=(n+1)P{N=n+1}E[N]=n+1rprn+1pn+1(1-p)r-n-1=n+1rpr!(r-1-n)!(n+1)!pn+1(1-p)r-1-n=(r-1)!(r-1-n)!n!pn(1-p)r-1-n

Thus, M-1 is a binomial random variable with parameters r-1,p.

Fixing p, let N(r) be a binomial random variable with parameters r and p, and let

Pr(k)=P{SN(r)=k}

Then, Corollary 3.6 yields

Pr(0)=(1-p)rPr(k)=rpk∑j=1kjαjPr-1(k-j),k>0

For instance, letting k equal 1, then 2, and then 3 gives

Pr(1)=rpα1(1-p)r-1Pr(2)=rp2[α1Pr-1(1)+2α2Pr-1(0)]=rp2(r-1)pα12(1-p)r-2+2α2(1-p)r-1Pr(3)=rp3α1Pr-1(2)+2α2Pr-1(1)+3α3Pr-1(0)=α1rp3(r-1)p2(r-2)pα12(1-p)r-3+2α2(1-p)r-2+2α2rp3(r-1)pα1(1-p)r-2+α3rp(1-p)r-1

3.7.3 A Compounding Distribution Related to the Negative Binomial

Suppose, for a fixed value of p,0<p<1, the compounding random variable N has a probability mass function

P{N=n}=n+r-1r-1pr(1-p)n,n=0,1,…

Such a random variable can be thought of as being the number of failures that occur before a total of r successes have been amassed when each trial is independently a success with probability p. (There will be n such failures if the rth success occurs on trial n+r. Consequently, N+r is a negative binomial random variable with parameters r and p.) Using that the mean of the negative binomial random variable N+r is E[N+r]=r/p, we see that E[N]=r1-pp.

Regard p as fixed, and call N an NB(r) random variable. The random variable M-1 has probability mass function

P{M-1=n}=(n+1)P{N=n+1}E[N]=(n+1)pr(1-p)n+rr-1pr(1-p)n+1=(n+r)!r!n!pr+1(1-p)n=n+rrpr+1(1-p)n

In other words, M-1 is an NB(r+1) random variable.

Letting, for an NB(r) random variable N,

Pr(k)=P{SN=k}

Corollary 3.6 yields

Pr(0)=prPr(k)=r(1-p)kp∑j=1kjαjPr+1(k-j),k>0

Thus,

Pr(1)=r(1-p)pα1Pr+1(0)=rpr(1-p)α1,Pr(2)=r(1-p)2pα1Pr+1(1)+2α2Pr+1(0)=r(1-p)2pα12(r+1)pr+1(1-p)+2α2pr+1,Pr(3)=r(1-p)3pα1Pr+1(2)+2α2Pr+1(1)+3α3Pr+1(0)

and so on.

Exercises

1. If X and Y are both discrete, show that ∑xpX∣Y(x∣y)=1 for all y such that pY(y)>0.

*2. Let X1 and X2 be independent geometric random variables having the same parameter p. Guess the value of

P{X1=i∣X1+X2=n}

Hint: Suppose a coin having probability p of coming up heads is continually flipped. If the second head occurs on flip number n, what is the conditional probability that the first head was on flip number i,i=1,…,n-1?
Verify your guess analytically.

3. The joint probability mass function of X and Y,p(x,y), is given by

p(1,1)=19,p(2,1)=13,p(3,1)=19,p(1,2)=19,p(2,2)=0,p(3,2)=118,p(1,3)=0,p(2,3)=16,p(3,3)=19

Compute E[X∣Y=i] for i=1,2,3.

4. In Exercise 3, are the random variables X and Y independent?

5. An urn contains three white, six red, and five black balls. Six of these balls are randomly selected from the urn. Let X and Y denote respectively the number of white and black balls selected. Compute the conditional probability mass function of X given that Y=3. Also compute E[X∣Y=1].

*6. Repeat Exercise 5 but under the assumption that when a ball is selected its color is noted, and it is then replaced in the urn before the next selection is made.

7. Suppose p(x,y,z), the joint probability mass function of the random variables X,Y, and Z, is given by

p(1,1,1)=18,p(2,1,1)=14,p(1,1,2)=18,p(2,1,2)=316,p(1,2,1)=116,p(2,2,1)=0,p(1,2,2)=0,p(2,2,2)=14

What is E[X∣Y=2]? What is E[X∣Y=2,Z=1]?

8. An unbiased die is successively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a six and a five. Find (a) E[X], (b) E[X∣Y=1], (c) E[X∣Y=5].

9. Show in the discrete case that if X and Y are independent, then

E[X∣Y=y]=E[X]for ally

10. Suppose X and Y are independent continuous random variables. Show that

E[X∣Y=y]=E[X]for ally

11. The joint density of X and Y is

f(x,y)=(y2-x2)8e-y,0<y<∞,-y⩽x⩽y

Show that E[X∣Y=y]=0.

12. The joint density of X and Y is given by

f(x,y)=e-x/ye-yy,0<x<∞,0<y<∞

Show E[X∣Y=y]=y.

*13. Let X be exponential with mean 1/λ; that is,

fX(x)=λe-λx,0<x<∞