1. ## Conditional probability

Hi can you help me to solve this problem? but without theory stochastic. I need to use transformations and probability theory. thank you

Let X1 and X2 be items of a random sample from a distribution with p.d.f. (density function) f(x)= 2x , 0<x<1 , zero elsewhere. Evaluate the conditional probability Pr(X1<X2|X1<2X2)

2. $\displaystyle P(X_{1}<X_{2}) = \frac{1}{2}$

$\displaystyle P(X_{1} < 2X_{2}) = \int^{1}_{0} P(X_{1}<2X_{2}|X_{1}=x)f(x) \ dx$

$\displaystyle = 2\int^{1}_{0} xP(x<2X_{2}) \ dx = 2 \int^{1}_{0} x P(x/2 < X_{2}) dx$

$\displaystyle = 2 \int^{1}_{0} x \Big(2\int^{1}_{x/2}x \ dx \Big) \ dx = \int^{1}_{0} (2x-\frac{x^{3}}{2}) \ dx$

$\displaystyle = x^{2} - \frac{x^{4}}{8} \Big|^{1}_{0} = 1-\frac{1}{8} = \frac{7}{8}$

so $\displaystyle P(X_{1}<X_{2}|X_{1}<2X_{2}) = \frac{P( X_{1}<X_{2}, X_{1}<2X_{2})}{P(X_{1}<2X_{2})} = \frac{P(X_{1}<X_{2})}{P(X_{1}<2X_{2})}$

$\displaystyle = \frac{1/2}{7/8} = \frac{4}{7}$

3. ## Probability

Hi can you help me to solve this problem? But using probability theory, but no stochastic theory. I need to use the transfomations in this exercise. Thank you

Let X1 and X2 be items of a random sample from a distribution with p.d.f. (density function) f(x)= 2x , 0<x<1 , zero elsewhere. Evaluate the conditional probability Pr(X1<X2|X1<2X2)