How to find the variance of $Y=3X-1$

Question

Here is the question:

Consider the probability density function $f(x)=2(x-1)$, for $0<x<1$, and $=0$ otherwise

Find $\sigma^2_y$ where $Y=3X-1$

I have the final answer with no steps and it is equal to $\frac{1}{2}$ but I can't seem to reach it.

Here is my attempt:

First, we need to find the expected value, we get $\int_0^1 (3x-1)2(1-x)dx=0$

For the variance we apply the general formula $\int_0^1 (x-\mu)^2\ f(x)dx$ with $\mu=0$ , and here comes the trouble, I tried doing $\int_0^1 (x)^2\ (3x-1)dx$ and it didn't give the desired answer
I also tried $\int_0^1 (x)^2\ 3[(2(1-x)]-1dx$ and it was also wrong (this one was just a random attempt, I do not think it is logically correct since it is more like $g(f(x))$, I have tried looking around the book for a formula to apply this with no luck, any help?

Answer 1

Simply use variance property that is

$$\mathbb{V}[aX+b]=a^2\mathbb{V}[X]$$

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there is an evident typo in the exercise and the correct one is

$$f_X(x)=2(1-x)$$

thus

$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]=\frac{3}{54}$$

and finally

$$\mathbb{V}[Y]=9\mathbb{V}[X]=\frac{1}{2}$$

Answer 2

Alert: The probability probability density function you gave is bad: it is negative over the given support. Please check that you have copied it correctly. I suspect that you instead meant $f(x)=2(1-x)\,\mathbf 1_{0<x<1}$

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and here comes the trouble, I tried doing $\int_0^1(x)^2 (3x−1)dx$ ...

Yes, that is trouble. Don't do that.

Instead you should do the same thing as you did earlier: substitute $3x-1$ for $y$ .

$$\begin{align}\mathsf E(Y)&=\int_\Bbb R y\,f(x)\,\mathrm d x&&\text{where }y=3x-1\\&=\int_0^1 (3x-1)\cdot 2(1-x)\,\mathrm d x\\&=2\int_0^1 -3x^2+4x-1\,\mathrm d x\\&=0\\[2ex]\mathsf {Var}(Y)&=\int_\Bbb R (y-\mathrm E(Y))^2\,f(x)\,\mathrm d x&&\text{where }y=3x-1\\&=\int_0^1((3x-1)-0)^2\cdot 2(1-x)\,\mathrm d x\\&=2\int_0^1 -9x^3+15x^2-7x+1\,\mathrm d x\\&=\dfrac 12\end{align}$$