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Properties of expectation - Discrete Probabilities

Properties of expectation

Lessons

Notes:
We can write out mean as an expected value:
μ=E[X]\mu=E[X]

And likewise for variance:
σ2=\sigma^2= Var(X)(X)

n: number of trials
x: number of success in n trials
p: probability of success in each trial

Binomial:
E[X]=npE[X]=np
Var(X)=np(1p)(X)=np(1-p)

Geometric:
E[X]=1pE[X]=\frac{1}{p}
Var(X)=1pp2(X)=\frac{1-p}{p^2}

Properties of Expectation:

\cdot E[X+a]=E[X]+aE[X+a]=E[X]+a
\cdot E[bX]=bE[X]E[bX]=bE[X]
\cdot E[X+Y]=E[X]+E[Y]E[X+Y]=E[X]+E[Y]

Or in full generality:
\cdot E[X1+X2++Xn]=E[X1]+E[X2]++E[Xn]E[X_1+X_2+ \cdots +X_n ]=E[X_1 ]+E[X_2 ]+ \cdots +E[X_n]

Properties of Variance:
\cdot Var[X+a]=[X+a]= Var[X][X]
\cdot Var[bX]=b2[bX]=b^2Var[X][X]
\cdot Var[X+Y]=[X+Y]= Var[X]+[X]+ Var[Y][Y] if X and Y are independent
  • 1.
  • 3.
    A certain car breaks down every 50 hours of driving time. If the car is driven for a total of 175 hours;
  • 4.
    Clara is trying to make the perfect teapot out of pottery. Each time she attempts to make the perfect teapot she will use a lump of clay and she will succeed with a probability of 0.20. Once she makes the perfect teapot she will stop potting.
  • 6.
    Suppose we have two independent random variable one with parameters E[X]=4E[X]=4 and Var(X)=3(X)=3, and the other with parameters E[Y]=9E[Y]=9 and Var(Y)=6(Y)=6.
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Properties of expectation

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