Linear combinations of random variables Flashcards

(6 cards)

1
Q

Mean of a function of a random variable 𝑬(𝑿) = Β΅

A

𝐸(π‘Žπ‘‹ + 𝑏) = π‘Ž Γ— 𝐸(𝑋) + 𝑏

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2
Q

Variance of a function of a random variable 𝑽𝒂𝒓(𝑿) = 𝝈² (3)

A

∴ π‘‰π‘Žπ‘Ÿ(π‘Žπ‘‹ + 𝑏) = π‘ŽΒ² Γ— π‘‰π‘Žπ‘Ÿ(𝑋)
∴ 𝑆𝑑(π‘Žπ‘‹ + 𝑏) = |π‘Ž| Γ— 𝑆𝑑(𝑋)

Note: π‘‰π‘Žπ‘Ÿ(𝑋) = 𝐸(𝑋²) – [E(X)]Β²

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3
Q

Sum and difference of independent random variables

A

For T = X + Y
E(T) = E(X + Y) = E(X) + E(Y)
Var(T) = Var(X + Y) = Var(X) + Var(Y)

For T = X - Y
E(T) = E(X - Y) = E(X) - E(Y)
Var(T) = Var(X - Y) = Var(X) + Var(Y)

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4
Q

Any random variables vs independent random variables

A

For any random variables:
E(aX + bY) = a x E(x) + b x E(Y)

For independent random variables:
Var(aX + bY) = aΒ² x Var(X) + bΒ² x Var(Y)

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5
Q

Linear functions and combinations of normally distributed random variables

A

Linear combination of independent normal variables are also normally distributed

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6
Q

The distribution of the sum of 2 independent Poisson variables

A

X~Po(Ξ»β‚“) and Y~Po(Ξ»α΅§) then,
E(X + Y) = E(X) + E(Y) = Ξ»β‚“ + Ξ»α΅§

Var(X + Y) = Var(X) + Var(Y) = Ξ»β‚“ + Ξ»α΅§

Mean and variance of X and Y are equal hence X + Y have a poisson distribution given that X and Y are independent

Note:
Linear combination of independent poisson variables of the for aX + bY can not have a poisson distribution

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