Mean and Variance of Continuous RV

Probability Level 2

Suppose a continuous random variable X X is given by the PDF:

f ( x ) = { 2 x x ∈ [ 0 , 1 ] 0 otherwise. f(x) = \begin{cases} 2x \quad & x \in [0,1] \\ 0 \quad & \text{otherwise.} \end{cases}

If the mean of X X is A A and the variance of X X is B B , what is A + B A+B ?

2 3 \frac23 13 18 \frac{13}{18} 4 9 \frac{4}{9} 1 18 \frac{1}{18}

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Matt DeCross by Matt DeCross , 27, USA

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1 solution

Sam Bealing
Apr 26, 2016

E ( X ) = ∫ 0 1 x × 2 x = ∫ 0 1 2 x 2 = [ 2 3 x 3 ] 0 1 = 2 3 E(X)= \int_{0}^{1} {x \times 2x}=\int_{0}^{1} 2x^2=\left [ \dfrac{2}{3}x^3 \right ]_{0}^1=\dfrac{2}{3}

E ( X 2 ) = ∫ 0 1 x 2 × 2 x = ∫ 0 1 2 x 3 = [ 1 2 x 4 ] 0 1 = 1 2 E(X^2)= \int_{0}^{1} {x^2 \times 2x}=\int_{0}^{1} 2x^3=\left [ \dfrac{1}{2} x^4 \right ]_{0}^1=\dfrac{1}{2}

V a r ( X ) = E ( X 2 ) − ( E ( X ) ) 2 = 1 2 − ( 2 3 ) 2 = 1 18 Var(X)=E(X^2)-(E(X))^2=\dfrac{1}{2}-(\frac{2}{3})^2=\dfrac{1}{18}

B = E ( X ) = 2 3 , A = V a r ( X ) = 1 18 ⇒ A + B = 13 18 B=E(X)=\dfrac{2}{3},A=Var(X)=\dfrac{1}{18} \Rightarrow A+B=\boxed{\dfrac{13}{18}}

4 Helpful 2 Interesting 0 Brilliant 2 Confused

Moderator note:

Simple standard approach.

you did calculation mistake in varience please check

Madhu V - 3 years, 4 months ago

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