about random variable and Binomial distribution

[KFC]

Hi there,
As many texts' discussion, we usually use a variable x for any value randomly picked. For a Bernoulli trials, i.e. each random variable x can either be successful or fail. If the probability of success if p and that of failure is q=1-p, then the expectation value of x would be

\langle x\rangle = x_s p + x_f(1-p)

where x_s is the value of success while x_f is the value of failure.

In many texts, it takes x_s=1 and x_f=0. Hence,

\langle x\rangle = x_s p + x_f(1-p) = p

I wonder why and from what point shall we define success and failure as x_s=1 and x_f=0? Why I can't say x_s=1 and x_f=-1 OR
x_s=0 and x_f=1? But it we change the valus of x_s and x_f, \langle x\rangle will definitely be changed!?

[mathman]

What you are suggesting is perfectly valid. Using 1 for success and 0 for failure is a convention to keep things simple. Changing to other values doesn't affect the ideas, only the arithmetic.

[KFC]

What you are suggesting is perfectly valid. Using 1 for success and 0 for failure is a convention to keep things simple. Changing to other values doesn't affect the ideas, only the arithmetic.

Thanks. But how? It is known that \langle x\rangle = p, but if we assume for example x_s=1 and x_f=-1, then

\langle x\rangle = x_s p + x_f(1-p) = p - (1-p) = 2p - 1

which is not consistent with \langle x\rangle = p

[statdad]

Thanks. But how? It is known that \langle x\rangle = p, but if we assume for example x_s=1 and x_f=-1, then

\langle x\rangle = x_s p + x_f(1-p) = p - (1-p) = 2p - 1

which is not consistent with \langle x\rangle = p

You get \langle x \rangle = p because of the current assignment of 1 and 0 to success and failure. Had the assignments been made some other way originally the expectation would be some other value.

The assignment isn't really arbitrary: in applications binomial rvs are used to count (record) the total number of successes that occur. Assigning -1 to indicate the occurrence of a failure works mathematically but it makes the application more difficult to deal with.

[bpet]

The standard Bernoulli variable is sufficiently general to represent any other combination of outcomes, e.g.

X = x_f + (x_s-x_f)B

where B is Bernoulli. As an affine function it's easy enough to calculate the mean and variance.

[KFC]

Thanks guys. All right, I get some points here, if we change the random variable, the average will change, just like we use a dice with 6 different values but ranged from 5 to 11, the average,of course, will be different from that ranged from 1 to 6. Is my logic right?

Now let consider a more general question on variance, it is easy to get a general expression in terms of x_s and x_f as follows

VARIANCE[X] = (x_s-x_f)^2pq

I understand that if we change the assignment of x_s and x_f, the VARIANCE will also changed by a factor (x_s-x_f)^2, but what's the significance of this factor (x_s-x_f)^2.Or I change my question to: any practical application in whichx_s\neq 0 and x_f\neq 1?

You get \langle x \rangle = p because of the current assignment of 1 and 0 to success and failure. Had the assignments been made some other way originally the expectation would be some other value.

The assignment isn't really arbitrary: in applications binomial rvs are used to count (record) the total number of successes that occur. Assigning -1 to indicate the occurrence of a failure works mathematically but it makes the application more difficult to deal with.