1. ## Uniform/ exponential dist.

I think these are the last two I can't get, from the four sections. I don't really even know where to start:

1.) Let X have an exponential dist. with mean, theta > 0. Show that

P(X > x + y | X > x) = P(X > y)

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2.) Let Y have a uniform dist. U (0, 1)

and let W=a +(b-a)Y a<b

a) Find the dist. function of W. [Hint: find P[a+ (b-a)Y < or =w]

b) How is W distributed?

2. 1.) Let X have an exponential dist. with mean, theta > 0. Show that

P(X > x + y | X > x) = P(X > y)
remember that
$\displaystyle P(X > x + y | X > x) = \frac{P(X>x+y)}{P(X>x)}$

P(X>a) =1-F(a) where F is the cdf of X.
This is where to start. By the sound of it I think you will be able to continue nicely.

2.) Let Y have a uniform dist. U (0, 1)

and let W=a +(b-a)Y a<b

a) Find the dist. function of W. [Hint: find P[a+ (b-a)Y < or =w]

b) How is W distributed?
Start by finding the distribution function of Y. Then note that the function W(Y) is strictly increasing. Can you continue?

3. Nope, I still don't understand. Sorry. I'm not sure what to do next.

4. Originally Posted by amor_vincit_omnia
I think these are the last two I can't get, from the four sections. I don't really even know where to start:

1.) Let X have an exponential dist. with mean, theta > 0. Show that

P(X > x + y | X > x) = P(X > y)

--------

2.) Let Y have a uniform dist. U (0, 1)

and let W=a +(b-a)Y a<b

a) Find the dist. function of W. [Hint: find P[a+ (b-a)Y < or =w]

b) How is W distributed?
1.) $\displaystyle f(x) = \frac{1}{\theta} \, e^{-\frac{x}{\theta}}, ~ x \geq 0$, and zero otherwise.

$\displaystyle \Pr(X > x + y | X > x) = \frac{ 1 - \int_{0}^{x+y} \frac{1}{\theta} \, e^{-\frac{t}{\theta}} \, dt }{1 - \int_{0}^{x} \frac{1}{\theta} \, e^{-\frac{t}{\theta}} \, dt } = \frac{ \theta - \int_{0}^{x+y}e^{-\frac{t}{\theta}} \, dt }{\theta - \int_{0}^{x}e^{-\frac{t}{\theta}} \, dt } = ......$

$\displaystyle \Pr(X > y) = 1 - \int_{0}^{y} \frac{1}{\theta} \, e^{-\frac{t}{\theta}} \, dt = .....$

You should get the same answer for both.

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2.) Many aproaches are possible. Here's one:

Consider the cdf of W: $\displaystyle F(w) = \Pr(W < w) = \Pr(a + (b-a)Y < w) = \Pr \left( Y < \frac{w - a}{b - a} \right) = \int_{0} ^{\frac{w-a}{b-a}} 1 \, dy = .....$, $\displaystyle a \leq w \leq b$ (since $\displaystyle 0 \leq y \leq 1$ ) and zero otherwise.

Then $\displaystyle f(w) = \frac{dF}{dw} = .......$.

Unsurprisingly, it's seen that W ~ U(a, b) ......

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# let y have a uniform distribution U(0,1) and let W=a (b-a)y find the cdf of w

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