# Mixed random variable distribution question

1. Oct 25, 2017

### King_Silver

1. The problem statement, all variables and given/known data
See attached image (See below)

2. Relevant equations
Differential equations.
And a combination of discrete & continuous distributions

3. The attempt at a solution
The Continous Distribution Function (CDF) is given in the question. So I differentiated it with respect to x piecewise (See below). That is what I got

I can also see that there is a uniform distribution between 0 and 1 but also there are discrete points that have assigned probability. For example; those points could be 0,1 and 2? maybe?
I want to be able to find the value of c while remembering the sum of the probabilities must add up and equal to 1.

How do I go about solving for c? this is where I am stuck. I cannot seem to figure out where to go from here. Thanks!

Question:
(When I differentiated it)

2. Oct 25, 2017

### andrewkirk

The calculation of the density function $f$ is not quite correct. The density is infinite at $x=0,2$ and also possibly at 1 (depending on the value of $c$), because the CDF jumps up discontinuously at those points. We say there is a 'discrete probability mass' at those infinite-density points, meaning there is a non-zero probability of X having exactly that value.

You need to use the additional fact that $E[X]=1$. Using the formulas you got for $f$, corrected to also include the points of discrete probability mass at 0 and 2, write an expression for $E[X]$ in terms of $c$. Equating that expression to 1 will give you an equation you can solve to find $c$. The expression will be a sum of integrals and isolated discrete terms.

There are two cases to consider here - (1) where there is a discrete probability mass at $x=1$ and (2) where there is not. I suggest trying case (1) first. If that does not give an acceptable solution, try case (2).

3. Oct 25, 2017

### King_Silver

Thank you for that. I figured it might have had something to do with the expectation formula. E[X] = 1 in terms of c. There are definitions for the expectation of continuous random variables and for discrete ones as well. For a mixed distribution you would need to mix both definitions together right?
I'm having a lot of trouble trying to figure out how to mix these two expectation formulas

4. Oct 25, 2017

### andrewkirk

Just integrate $xf(x)$ over the domain excluding the discrete mass points, and add the sum across each of those points of (probability of landing on the point) x (value of x at that point).

5. Oct 25, 2017

### King_Silver

This is what I did but I don't think I'm going in a correct direction. C = constant whereas c = the value i'm looking for.

6. Oct 25, 2017

### Ray Vickson

For a non-negative random variable $X$, whether continuous, discrete or mixed, if $F(x) = P(X \leq x)$ is the CDF and $G(x) = P(X > x) = 1 - F(x)$ is the complementary CDF, then we have that
$$EX = \int_0^{\infty} G(x) \, dx$$
You can do the integral in terms of $c$ and so get an equation for $c$.

7. Oct 25, 2017

### King_Silver

I got a c value of 1/3 by doing this (1 = c/2 + ½ -c + 2/3 )
Then for part (i): Probability = 1/6th or 16.67%
(ii) Probability = 2/3rds or 66.67%
(iii) Probability = 1/3rd or 33.3%
(iv) Probability = 1/2 or 50%

I think I've got it now providing those are the right answers!

8. Oct 25, 2017

### Ray Vickson

The first two are; I have not checked the other two.

However, in English, "1/6" is already pronounced as "one-sixth", so when you write 1/6th you are really writing "one-sixth-th". Thus, the The first two answers are just 1/6 and 2/3 (not "two-thirds-irds").

Last edited: Oct 25, 2017