Question #236314

Let X be a Bernoulli random variable (Hint: Special case of Binomial Distribution)

a. Compute E(X2)

b. Show that V(X) is p(1-p)

c. Compute E(X79)


1
Expert's answer
2021-09-14T00:58:08-0400

Solution:

XBernoulli(p),0p1X \sim \operatorname{Bernoulli}(p), 0 \leq p \leq 1

So, P(X=x)={1p if x=0p if x=1P(X=x)= \begin{cases}1-p & \text { if } x=0 \\ p & \text { if } x=1\end{cases}  

Now the expected value of X:

 E(X)=0P(X=0)+1P(X=1)=0×(1p)+1×p=p\begin{aligned} E(X) &=0 \cdot P(X=0)+1 \cdot P(X=1) \\ &=0 \times(1-p)+1 \times p \\ &=p \end{aligned}  

(a):

 E(X2)=02P(X=0)+12P(X=1)=0×(1p)+1×p=p\begin{aligned} E\left(X^{2}\right) &=0^{2} \cdot P(X=0)+1^{2} \cdot P(X=1) \\ &=0 \times(1-p)+1 \times p \\ &=p \end{aligned}  

(b):

 Var(X)=E(X2)E2(X)=pp2=p(1p)\begin{aligned} \operatorname{Var}(X) &=E\left(X^{2}\right)-E^{2}(X) \\ &=p-p^{2} \\ &=p(1-p) \end{aligned}

(c):

E(X79)=079P(X=0)+179P(X=1)=0×(1p)+1×p=p\begin{aligned} E\left(X^{79}\right) &=0^{79} \cdot P(X=0)+1^{79} \cdot P(X=1) \\ &=0 \times(1-p)+1 \times p \\ &=p \end{aligned}


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Comments

Neil
14.09.21, 13:50

Thanks for the excellent work and services !!

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