# Math Help - Checking a continous random variable distrubtion

1. ## Checking a continous random variable distrubtion

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

2. 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.

3. 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

4. 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.

5. 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.