Checking a continous random variable distrubtion

**thomas49th** · May 31st 2009, 04:10 AM

apparently, according to a paper mark scheme

$ f(y) = \left\{ \begin{array}{l l} \frac{1}{9} (2y^{3} + y), & 1 \leq y \leq 2 \\ & \\ 0 & \text{otherwise} \end{array} \right. $

Anyway im interested in $\frac{1}{9}(2y^{3}+y)$

is a valid probability density function
But if you put the upper limit 2 in you get 2 out, when probability ranges between 0 and 1.

Thanks

**mr fantastic** · May 31st 2009, 04:22 AM

Originally Posted by thomas49th

apparently, according to a paper mark scheme

$ f(y) = \left\{ \begin{array}{l l} \frac{1}{9} (2y^{3} + y), & 1 \leq y \leq 2 \\ & \\ 0 & \text{otherwise} \end{array} \right. $

Anyway im interested in $\frac{1}{9}(2y^{3}+y)$

is a valid probability density function
But if you put the upper limit 2 in you get 2 out, when probability ranges between 0 and 1.

Thanks

$\frac{1}{9} \int_1^2 (2y^{3} + y) \, dy = 1$ so I don't see what the problem is.

**thomas49th** · May 31st 2009, 04:45 AM

I thought a PDF tells you what the probability is for a certain value you put in.
My PDF says that between 1 and 2 the probabilty of getting a value inbetween this is

$ \frac{1}{9}(2y^{3}+y) $

where y is the value you want to find the probability of getting it

BUT if you put a 2 in you get a probability greater than 1, which cant be

Right?

Thanks

**Moo** · May 31st 2009, 04:49 AM

Hello,

Originally Posted by thomas49th

I thought a PDF tells you what the probability is for a certain value you put in.
My PDF says that between 1 and 2 the probabilty of getting a value inbetween this is

$ \frac{1}{9}(2y^{3}+y) $

where y is the value you want to find the probability of getting it

BUT if you put a 2 in you get a probability greater than 1, which cant be

Right?

Thanks

I think you're misunderstanding the role of the pdf

Two things :
- the probability that a continuous random variable takes a specific value is 0.
- if X has the pdf f, then :
$\mathbb{P}(a\leq X\leq b)=\int_a^b f(x)~dx$ , and not something like $f(b)-f(a)$

A third thing... :
A function is a pdf if : it is a positive function, and its integral over $\mathbb{R}$ is 1.

**mr fantastic** · May 31st 2009, 04:51 AM

Originally Posted by thomas49th

I thought a PDF tells you what the probability is for a certain value you put in.
My PDF says that between 1 and 2 the probabilty of getting a value inbetween this is

$ \frac{1}{9}(2y^{3}+y) $

where y is the value

BUT if you put a 2 in you get a probability greater than 1, which cant be

Right?

Thanks

You have many wrong ideas.

1. If Y is a continuous random variable, then Pr(Y = a) = 0. It only makes sense to talk about the probabilty that Y lies between two values. You do not calculate probabilities by substituting values of y into the pdf.

2. For $y \geq 1$ , $\Pr(Y < y) = \frac{1}{9} \int_1^y (2t^{3} + t) \, dt$ .

You are strongly advised to extensively review all you have learned on continuous random variables and probability density functions.

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