c.d.f. of a random variable, given pdf

**Jame** · March 21st 2011, 06:53 PM

pdf is given by

$f(x) = 2-4|x| if -1/2 < x < 1/2$ and $0$ otherwise

I cant do the integral of $2-4|x|$ so I am going to make 2 cases, $x < 0$ and $x > 0.$

$If x < 0 , f(x) = 2+4x$ for $-1/2 < x < 0$

So we look at $\int_{x}^{0} 2+4t \text{d}t = -2x(x+1)$

$If x > 0 , f(x) = 2x-4x$ for $0 < x < 1/2$

So we look at $\int_{0}^{x} 2-4t \text{d}t = -2x(x-1)$

So I figure the cdf should be

$F_X(x) =$

$0$ if $x \leq -1/2$

$-2x(x+1)$ if $-1/2 < x < 0$

$-2x(x-1) + \int_{-1/2}^{0} 2+4t \text{d}t = -2x(x-1) + 1/2$
if $0 \leq x < 1/2$

$1$ if $x \geq 1/2$

So i think is make sense to put $0$ with the positive case. Since the adding of the $1/2$ captures all the probabilities for negative values. However I think the function for $-1/2 < x < 0$ is wrong. Lower numbers are giving me higher probabilities than numbers close to $0$ . So something must be wrong.

**mr fantastic** · March 21st 2011, 07:52 PM

Originally Posted by Jame

pdf is given by

$f(x) = 2-4|x| if -1/2 < x < 1/2$ and $0$ otherwise

I cant do the integral of $2-4|x|$ so I am going to make 2 cases, $x < 0$ and $x > 0.$

$If x < 0 , f(x) = 2+4x$ for $-1/2 < x < 0$

So we look at $\int_{x}^{0} 2+4t \text{d}t = -2x(x+1)$

$If x > 0 , f(x) = 2x-4x$ for $0 < x < 1/2$

So we look at $\int_{0}^{x} 2-4t \text{d}t = -2x(x-1)$

So I figure the cdf should be

$F_X(x) =$

$0$ if $x \leq -1/2$

$-2x(x+1)$ if $-1/2 < x < 0$

$-2x(x-1) + \int_{-1/2}^{0} 2+4t \text{d}t = -2x(x-1) + 1/2$
if $0 \leq x < 1/2$

$1$ if $x \geq 1/2$

So i think is make sense to put $0$ with the positive case. Since the adding of the $1/2$ captures all the probabilities for negative values. However I think the function for $-1/2 < x < 0$ is wrong. Lower numbers are giving me higher probabilities than numbers close to $0$ . So something must be wrong.

It should be apparent that:

Case 1: $x < -\frac{1}{2}$ . $\displaystyle F(x) = 0$ .

Case 2: $-\frac{1}{2} \leq x \leq 0$ . $\displaystyle F(x) = \int_{-1/2}^{x} (2 + 4t) \, dt$ .

Case 3: $0 < x \leq \frac{1}{2}$ . $\displaystyle F(x) = \frac{1}{2} + \int_{0}^{x} (2 - 4t) \, dt$ .

Case 4: $x > \frac{1}{2}$ . $\displaystyle F(x) = 1$ .

**Jame** · March 23rd 2011, 12:48 PM

I have another question about this random variable.

I got that the expected value,

$E[X] = \int_{-1/2}^{0} x(2+4x) \; \text{d}x + \int_{0}^{1/2} x(2-4x) \; \text{d}x = 0$

Can I not break up the function to calculate the expected value?

**mr fantastic** · March 23rd 2011, 06:43 PM

Originally Posted by Jame

I have another question about this random variable.

I got that the expected value,

$E[X] = \int_{-1/2}^{0} x(2+4x) \; \text{d}x + \int_{0}^{1/2} x(2-4x) \; \text{d}x = 0$

Can I not break up the function to calculate the expected value?

Yes you can, which is what you appear to have done. But I would have used the fact that f(x) is even and hence xf(x) is odd to get $\displaystyle \int_{-1/2}^{1/2} x f(x) \, dx = 0$ .

Math Help - c.d.f. of a random variable, given pdf

LinkBack

Thread Tools

Display

c.d.f. of a random variable, given pdf

Similar Math Help Forum Discussions

Variance of a Poisson random variable whose parameter is also a random variable

exponential random variable with a random mean?

Finding the marginal distribution of a random variable w/ a random variable parameter

What's the difference between discrete random variables and continous random variable

Mean of random variable

Search Tags