# Find expected value with pdf and LOTUS

I am currently trying to solve a problem and can't figure it out. I have done this before, but I can't remember all of the details and can't find a reference example.

Let's say I have a pdf

$$f(x)=\frac{x^2}{16}\,\,\text{ for }-2\le x\le 2\,;\,0\text{ otherwise }$$

and I want to find the expected value for $$Y = X^2$$.

With LOTUS I would do the following

$$E[X^2]=\int_{-\infty}^\infty x^2f_X(x)\,dx$$

So in this specific case, I would calculate

$$E[Y]=E[X^2]=\int_{-2}^2 x^2\cdot\frac{x^2}{16}\,dx=\int_{-2}^2 \frac1{16}x^4\,dx=\left(\frac1{80}2^5-\frac1{80}(-2)^5\right)$$

However, there are two things I am confused with and I can't remember or find a good example:

1. I do have the feeling I need to transform the boundaries. Maybe I am wrong...

2. If the the boundaries go from negative to positive, I have the feeling that the last term $$\left(\frac1{80}2^5-\frac1{80}(-2)^5\right)$$ is wrong. The closest thing I could find was Find expected value using CDF but maybe you could shed some light on it.

Thanks for the help!

• Use MathJax for formatting math. – StubbornAtom Feb 2 at 14:54
• LOTUS seems irrelevant here: you aren't asking about the expectation of a random variable, but only for the value of a ratio of integrals involving a density function. – whuber Feb 2 at 16:55

Your PDF doesn't integrate to $$1$$, so you need a suitable scalar in front of it, e.g. $$3/16$$ instead of $$1/16$$. But, let's assume it's valid.
1. You don't need to transform boundaries because the integration is in terms of $$x$$.
2. The last term is also correct because both your boundaries and the integral $$x^2f(x)$$ is calculated correctly.