# Probability density function, cumulative function.

1. Oct 16, 2011

### Bassalisk

1. The problem statement, all variables and given/known data
Random voltage is defined with its probability density function:
$$p_{\xi}(v)=2,25u(v+2)e^{-3(v+2)}+k\delta (v-2)$$

u-Heaviside step function

a) Find constant k.
b) What is the probability of a random variable to take value of 2.
c) Find the cumulative distribution function.

2. Relevant equations
$\int_{-\infty}^{+\infty}p_{\xi}(x)dx=1$
3. The attempt at a solution
I got that the constant k is 0,25. (if somebody would check that for me)

Now for the b), I am not sure. I know that when you have continuous distribution, probability that the function has 1 specific value is 0.

But this delta function at that point is confusing me. Does it change things? Should the probability be 0.25?

And I have troubles finding cumulative distribution. Help me out here.

Last edited: Oct 16, 2011
2. Oct 16, 2011

### Ray Vickson

This is an example of a "mixed" random variable that has both a continuous and discrete component. Other (perhaps more familiar) examples are lifetime distributions of products, which may have a nonzero probability at lifetime = 0, because the product may be non-functional right out of the box, but if it is functional when new, then its remaining life has a continuous distribution. Anyway, that is the role of the delta function: it describes a point-mass for probability at v = 2. (Are you sure you should have delta(v-2) rather than delta(v+2)? It seems strange to have a point mass in the middle of a continuous part.)

So, in your case, Pr{V = 2} = k, which is not zero.

The cumulative distribution is just F(x) = Pr(V <= x} = integral(p(v), v = -infinity..x), which ought to be easy for you to do. In this case F(x) is not continuous at x = 2; maybe that is what was confusing you.

RGV

3. Oct 16, 2011

### Bassalisk

Thank you for your reply. Im sure that the δ is at 2. We usually have that, unusual things :).