Proving Binomial Distribution Expected Value & Variance

[Yulia_sch]

hello,

i need to prive that for a binomial r.v X E[X]=NP and VAR(X)=NP(1-P).

I tried to prove it using the deffinition of expectation:

E[x]=$$\sum xi \stackrel{N}{i} p^{i}(1-p)^{n-i}$$

now what?

thanks...

 

[mathman]

hello,

i need to prive that for a binomial r.v X E[X]=NP and VAR(X)=NP(1-P).

I tried to prove it using the deffinition of expectation:

E[x]=$$\sum xi \stackrel{N}{i} p^{i}(1-p)^{n-i}$$

now what?

thanks...

E[x]=$$\sum xi \stackrel{N}{i} p^{i}(1-p)^{n-i}$$ is incorrect. Should be:
E[x]=$$\sum i \stackrel{N}{i} p^{i}(1-p)^{n-i}$$

For the second moment replace i by i².

 

[statdad]

One more comment: to help with the variance, instead of calculating

$$E[X] = \sum i^2 \binom{N}{i} p^i (1-p)^{n-i}$$

calculating

$$E[X (X-1)] = \sum i(i-1) \binom{N}{i} p^i (1-p)^{n-i}$$

will make the algebra required to work with the summation simpler.

Since

$$E[X(X-1)] = E[X^2] - E[X]$$

this, and the mean, will allow you to find the variance.