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Given the function:

$$ f(x) = \begin{cases} 0 & x<0, \\ a(b-x) & 0\le x\le b, \\ 0 & b

where $a, b ∈ (0, ∞)$. We also know that $\operatorname{E}(X) = 1$. I need to find out $a,b$ and the distribution function.

I made an equation system using $\operatorname{E}(X)=1$ and $F(X)=1$ from which I got $a = 2/9$ and $b = 3$, but I'm not really sure if my method is correct and I'm a bit lost on how to calculate the distribution function. Any help is greatly appreciated! :)

$$\operatorname{E}(X)=\int_0^b y f(y) \,dy,$$ This gave me $ab^3=6$.

$$F(X)=\int_0^b f(y) \, dy$$ This gave me $ab^2=2$.

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    Hi Kamil Pop, and welcome to MSE. Can you post your method so we can take a look at it and give you appropriate feedback? – 2017-02-12
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    Hi Adam! Edited the post for more clarity. – 2017-02-12

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You need for your density function to integrate to $1.$ According to your solution you have: $$\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^0 0\,dx + \int_0^3 \frac{2}{9}(3-x)\,dx + \int_3^\infty 0\,dx = \int_0^3 \frac{2}{9}(3-x)\,dx = 1,$$ which is correct. Then to get the mean, you need to find. $$E(X) = \int_0^3 x \cdot \frac{2}{9}(3-x)\,dx$$ You have a random variable with support $(0,3)$ so the mean cannot possibly be $6.$ Try the integration again.

Your answer for the CDF is in three parts: $F(x) = 0,$ for $x \in (\infty,0);\; F(x) = 1,$ for $x \in (3, \infty);$ and $F(x) = \int_0^x f(t)\,dt$ for $x \in [0, 3].$ You still need to evaluate the last integral. Make sure that you get $F(0) = 0$ and $F(3) = 1.$

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    Yup, all good! This helped a lot. Cheers! :) – 2017-02-12