probability distribution - quick question

**hunkydory19** · October 29th 2008, 08:11 AM

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

P(Y) = exp(-Y)

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

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

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

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

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.

Thanks in advance!

**mr fantastic** · October 29th 2008, 10:08 PM

Originally Posted by hunkydory19

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

P(Y) = exp(-Y)

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

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

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

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

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.

Thanks in advance!

Your pdf for U is almost correct. It's

$P(U) = \frac{1}{2} exp \frac{5 - U}{2}$ for ${\color{red}U \in [5..\infty)}$ and zero elsewhere.

Actually, I'd use the notation $f(u) = \frac{1}{2} exp \frac{5 - u}{2}$ for $u \geq 5$ and zero elsewhere.

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