# Thread: probability distribution - quick question

1. ## probability distribution - quick question

A variable $\displaystyle Y \in [0..\infty)$ has probability density

P(Y) = exp(-Y)

I'm trying to find the probability distribution of U = 2Y + 5, so I have done:

$\displaystyle P_y(y) = P_x(h(y))\frac{dh}{dy}$

$\displaystyle Y = h(U) = \frac{U - 5}{2} ; \frac{dh}{du} = \frac{1}{2}$

So $\displaystyle P(U) = \frac{1}{2} exp \frac{5 - U}{2}$

Could anyone please tell me if this is right or are probability density and probability distribution different things? I've looked up the defintions but I'm still confused.

2. Originally Posted by hunkydory19
A variable $\displaystyle Y \in [0..\infty)$ has probability density

P(Y) = exp(-Y)

I'm trying to find the probability distribution of U = 2Y + 5, so I have done:

$\displaystyle P_y(y) = P_x(h(y))\frac{dh}{dy}$

$\displaystyle Y = h(U) = \frac{U - 5}{2} ; \frac{dh}{du} = \frac{1}{2}$

So $\displaystyle P(U) = \frac{1}{2} exp \frac{5 - U}{2}$

Could anyone please tell me if this is right or are probability density and probability distribution different things? I've looked up the defintions but I'm still confused.

$\displaystyle P(U) = \frac{1}{2} exp \frac{5 - U}{2}$ for $\displaystyle {\color{red}U \in [5..\infty)}$ and zero elsewhere.
Actually, I'd use the notation $\displaystyle f(u) = \frac{1}{2} exp \frac{5 - u}{2}$ for $\displaystyle u \geq 5$ and zero elsewhere.