About random variable and Binomial distribution

  • Thread starter KFC
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  • #1
KFC
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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

[tex]\langle x\rangle = x_s p + x_f(1-p)[/tex]

where [tex]x_s[/tex] is the value of success while [tex]x_f [/tex] is the value of failure.

In many texts, it takes [tex]x_s=1[/tex] and [tex]x_f=0[/tex]. Hence,

[tex]\langle x\rangle = x_s p + x_f(1-p) = p[/tex]

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

Answers and Replies

  • #2
mathman
Science Advisor
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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.
 
  • #3
KFC
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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 [tex]\langle x\rangle = p[/tex], but if we assume for example [tex]x_s=1[/tex] and [tex]x_f=-1[/tex], then

[tex]\langle x\rangle = x_s p + x_f(1-p) = p - (1-p) = 2p - 1[/tex]

which is not consistent with [tex]\langle x\rangle = p[/tex]
 
  • #4
statdad
Homework Helper
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Thanks. But how? It is known that [tex]\langle x\rangle = p[/tex], but if we assume for example [tex]x_s=1[/tex] and [tex]x_f=-1[/tex], then

[tex]\langle x\rangle = x_s p + x_f(1-p) = p - (1-p) = 2p - 1[/tex]

which is not consistent with [tex]\langle x\rangle = p[/tex]
You get [tex] \langle x \rangle = p [/tex] 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.
 
  • #5
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The standard Bernoulli variable is sufficiently general to represent any other combination of outcomes, e.g.

[tex]X = x_f + (x_s-x_f)B[/tex]

where B is Bernoulli. As an affine function it's easy enough to calculate the mean and variance.
 
  • #6
KFC
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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 [tex]x_s[/tex] and [tex]x_f[/tex] as follows

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

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

You get [tex] \langle x \rangle = p [/tex] 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.
 

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