# About random variable and Binomial distribution

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!?

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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.

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 $$\langle x\rangle = p$$

Homework Helper
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.

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.

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 which$$x_s\neq 0$$ 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.