Thread: Mixing a uniform distribution with a normal distribution

Here is my Problem.

Let X~Uniform(0,2) and Y~Normal(0,1) be independent random variables. Let

Z ={1 if X > Y
{2 if X <Y

Find E[Z] and V ar[Z]. You may leave your answers in terms of denite integrals.

My big question is, how can we join a uniform and a normal distribution on the same probability density function.

Thank you

Originally Posted by solvj

Here is my Problem.

Let X~Uniform(0,2) and Y~Normal(0,1) be independent random variables. Let

Z ={1 if X > Y
{2 if X <Y

Find E[Z] and V ar[Z]. You may leave your answers in terms of denite integrals.

My big question is, how can we join a uniform and a normal distribution on the same probability density function.

Thank you

If you know you can find , which is the conditional probability that . Then integrate this multiplied by the pdf of to get the unconditional probability that .

CB

Originally Posted by solvj

Here is my Problem.

Let X~Uniform(0,2) and Y~Normal(0,1) be independent random variables. Let

Z ={1 if X > Y
{2 if X <Y

Find E[Z] and V ar[Z]. You may leave your answers in terms of denite integrals.

My big question is, how can we join a uniform and a normal distribution on the same probability density function.

Thank you

Some cautions have to be taken because X and Y are continuos variables and Z is a discrete variable. The most sure way is to valuate the probability...

(1)

The first step...

(2)

(3)

Now other steps are...

a) use (2) and (3) to obtain (1)...

b) compute the derivative of (1) respect to ...

c) compute...

(4)

d) use the result c) to compute , , ...

All these steps are left to You...

Kind regards

So if I understand quite good, I should get P(x>Y)=P(Z=1|X=x)

I need to find P(X>Y)=P(Z=1)=integral from 0 to 2 (P(Z=1|X=x)*P(X=x)dx)
But what is P(x>Y)??

Thanks to both of you, my problem is solved.