Poisson distribution maths question help?

Question

I've tried looking online, past papers, in my book but can't find help for it.

The random variable X has a Poisson distribution with mean 4. The random variable Y is defined by

y=4X+1

Find the mean and standard deviation of Y.

Answer 1

The Poisson distribution with average rate lambda has a mean equal to lambda and standard deviation of sqrt(lambda). So, in order to figure this out, you need to remember that if

Y = aX + b, then

mean of Y = a(mean of X) + b and

sd of Y = a(sd of X), since adding a constant doesn't change how spread out a random variable is.

So,

the mean of Y is 4(4) + 1 = 17, and

the sd of Y = 4(sqrt(4)) = 8.

Answer 2

Its been a while since I've done this). You would need to do a change of variables; find the form of the function of the poisson, then replace x with (y-1)/4 (dont forget to multiply by the determinant of the Jacobian matrix). Doing this you can rigorously determine the exact answer.

Or, you can think that since y = 4x+1 is a linear combination then the mean would be similarly transformed. if y = AX+B then the mean is A(E(X))+B and the variance is A*A*(V(X)).

Answer 3

Since X has Poisson distribution it means:

P(X) = P[X = k] = exp(-a)*a^k/k! k = 0,1,2,3,...

From the above follows [ave(x) means average of X]:

ave(X) = sum(X*P(X)) = a = 4.
var(X) = ave(X^2) - ave(X)^2 = a = 4,
where ave(X^2) = sum((X^2)*P(X)).
It follows: ave(X^2) = 20.

Now you have the function Y of random variable X that is Y = 4X + 1.

ave(Y) = sum(Y*P(X))
= sum((4X+1)*P(X))
= 4*sum(X*P(X)) + sum(P(X))
= 4*ave(X) + 1 = 17.

st_dev(Y)^2 = var(Y) = ave(Y^2) - ave(Y)^2.

ave(Y^2) = sum((4X+1)^2*P(X))
= sum((16*X^2 + 8*X + 1)*P(X))
= 16*ave(X^2) + 8*ave(X) + 1
= 16*20 + 8*4 + 1= 353.

Therefore var(Y) = 353 - 17^2 = 64 and st_dev(Y) = 8.