# Thread: P(x ≥ 0), e[y], v[y]

1. ## P(x ≥ 0), e[y], v[y]

If continuous random variable X follow distribution
, P(X ≥ 0), E[Y], V[Y]

I calculate C is:
1=∫[-∞,+∞]f(x)dx
=∫[-1,+2]cx^2dx
=9c/3,
C =1/3
but no idea how to calculate P(X ≥ 0), E[Y], V[Y]
anyone can help?

2. ## Re: P(x ≥ 0), e[y], v[y]

I believe:

$$P(X\ge 0) = \int_0^\infty f(x)dx =\int_0^2 \dfrac{x^2}{3}dx = \dfrac{8}{9}$$

$$E[Y] = E[X^2] = \int_{-\infty}^\infty x^2f(x)dx = \int_{-1}^2 \dfrac{x^4}{3}dx = \left. \dfrac{x^5}{15}\right|_{-1}^2 = \dfrac{11}{5}$$

$$V[Y] = E[Y^2] - (E[Y])^2 = E[X^4] - (E[Y])^2 = \int_{-1}^2 \dfrac{x^6}{3}dx-\dfrac{121}{25} = \dfrac{43}{7}-\dfrac{121}{25} = \dfrac{228}{175}$$

I could be wrong, but that is how I would do it.

3. ## Re: P(x ≥ 0), e[y], v[y]

I think you are correct, Thanks!