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Exercise :

Let $X$ be a random variable with probability density function :

$$f(x) = ax + bx^2, 0<x<1, a,b \in \mathbb R$$

(a) If $E[X]=0.6$, find the distribution function $F$ of $X$ and then calculate the value of the distribution function $F$ at $0.5$ .

(b) Find the probability $P(X<1/2 | X<3/4)$ .

Attempt :

I know you can calculate the probability distribution function by :

$$F(x)=\int_{-\infty}^t f(x)dt = \int_{-\infty}^t (ax + bx^2)dx $$

but I do not know the integration "ends" I should take for both the previous and following integral.

Probably I can use $E[X]$ for :

$$E[X] = \int_{-\infty}^{\infty} xf(x) dx$$

For (b), I do not know at all how to handle it.

Any thorough solution will be very much appreciated because I want to understand and grasp on how to solve similar exercises and problems.

Thanks a lot for your time !

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    $\begingroup$ You are supposed to solve the system $$\int_0^1f(x)dx=1\qquad\int_0^1xf(x)dx=0.6$$ $\endgroup$ – Did Jun 5 '17 at 20:39
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First we need to find $a$ and $b$. This is possible because we have two equations, from the definition of a probability density function and the definition of expected value: $$\int_0^1 f(x) dx = 1$$ and $$\int_0^1 xf(x) dx = E[X].$$ Solving these integrals gives two linear equations in $a$ and $b$. From there we can find values for $a$ and $b$.

For part $(a)$, the correct formula is \begin{align*} F(t) &= \int_{-\infty}^t f(x) \;dx \\ &= \int_0^t ax+bx^2 \;dx \end{align*} since $f(x) = 0$ outside the interval $[0,1]$.

For part $(b)$, by the definition of conditional probability we have \begin{align*} P(X < 1/2 \mid X < 3/4) &= \frac{P(X < 1/2 \land X < 3/4)}{P(X < 3/4)} \\ &= \frac{P(X < 1/2)}{P(X < 3/4)} \\ &= \frac{F(1/2)}{F(3/4)}. \end{align*}

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