Poisson distribution/S1 question?

Question

The random variable X has a Poisson distribution with mean 4. The random variable Y is defined by y=2X+8

(a) Show that the mean and variance of Y are equal (I've worked them both out to be 16, is this correct?)

(b) John thinks that, because of this, Y has a Poisson distribution. Explain briefly why this is not the case.

Answer 1

If X has a Poisson distribution
P(X,4) = (e^-4)(4^X)/X!
shown below.
If Y = 2X + 8 it will have the distribution shown at the right.
X , . . . . . f . . . . . . , Y , . . . . . . f
0 , 0.018315639 , 8 , 0.832393986
1 , 0.073262556 , 10 , 0.147981153
2 , 0.146525111 , 12 , 0.017937109
3 , 0.195366815 , 14 , 0.001576889
4 , 0.195366815 , 16 , 0.000105126
5 , 0.156293452 , 18 , 5.49678E-06
6 , 0.104195635 , 20 , 2.31443E-07
7 , 0.059540363 , 22 , 8.01535E-09
8 , 0.029770181 , 24 , 2.32329E-10
9 , 0.013231192 , 26 , 5.71887E-12
10 , 0.00529247 , 28 , 1.21034E-13
11 , 0.00192453 , 30 , 2.22592E-15
12 , 0.00064151 , 32 , 3.59019E-17
13 , 0.00019738 , 34 , 5.1197E-19
14 , 5.63960E-05 , 36 , 6.50121E-21
15 , 1.50391E-05 , 38 , 7.39825E-23
16 , 3.75978E-06 , 40 , 7.58794E-25
17 , 8.84654E-07
18 , 1.96590E-07
19 , 4.13873E-08
20 , 8.27746E-09
21 , 1.57666E-09
22 , 2.86665E-10
23 , 4.98549E-11
24 , 8.30914E-12
25 , 1.32946E-12
26 , 2.04533E-13
27 , 3.03011E-14
28 , 4.32874E-15
29 , 5.97067E-16
30 , 7.96089E-17
31 , 1.02721E-17
32 , 1.28401E-18
33 , 1.55638E-19
34 , 1.83104E-20
35 , 2.09261E-21
36 , 2.32513E-22
37 , 2.51365E-23
38 , 2.64595E-24
39 , 2.71379E-25
40 , 2.71379E-26

(a) I get a mean of 8.378070945and a variance of 8.473667172 for Y.

(b) Y does not have a Poisson distribution because of the offset and the missing odd numbers.

If y = 2x + 8, discrete and uniformly distributed, the mean and the variance are equal with a value of 16 only at x = 8

Answer 2

a) E(Y) = E(2X + 8) = 2E(X) + E(8) = 2*4 + 8 = 16
Variance (X) is written as V(X)
V(Y) = V(2X+8) = V(2X) + V(8) = 4V(X) + 0 = 4*4 = 16

b) Because when E(Y)=V(Y) that does not imply Poisson. The sum of Poisson random variables is Binomial, I think.