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Exercise 11.6 · Q13

Q.The random variable XX has the probability density function
[!FORMULA] f(x)={ax+b0<x<10otherwisef(x)=\begin{cases}ax+b & 0<x<1\\ 0 & \text{otherwise}\end{cases}
and E(X)=712E(X)=\dfrac7{12}. Then aa and bb are respectively

(1) 11 and 12\dfrac12
(2) 12\dfrac12 and 11
(3) 22 and 11
(4) 11 and 22.
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Normalisation gives one linear relation between aa and bb; the given mean gives a second; solve the pair simultaneously.

Step 1. Normalisation. ∫01(ax+b) dx=a2+b=1\displaystyle\int_0^1(ax+b)\,dx=\dfrac a2+b=1. …(I)

Step 2. Mean condition. E(X)=∫01x(ax+b) dx=∫01(ax2+bx) dx=a3+b2=712E(X)=\displaystyle\int_0^1x(ax+b)\,dx=\int_0^1(ax^2+bx)\,dx=\dfrac a3+\dfrac b2=\dfrac7{12}. …(II)

Step 3. Solve (I) for bb. b=1−a2b=1-\dfrac a2.

Step 4. Substitute into (II). a3+12 ⁣(1−a2)=712 ⇒ a3+12−a4=712\dfrac a3+\dfrac12\!\left(1-\dfrac a2\right)=\dfrac7{12}\ \Rightarrow\ \dfrac a3+\dfrac12-\dfrac a4=\dfrac7{12}. …

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