given the following CDF, find the expected value

Question

I got stuck at the middle of the question. would appreciate your help. first of all, given the CDF as follows, I had to find parameters $a$ and $b$ such that the CDF is a function of a continuous random variable.

So as I know, $F_x(t<=4)=1$ then $ 0.4t+a=1$ while $t=4$. then $a=-0.6$ and $b=0.2$.

after that, I had to find the density function value for x=3. and that would be the derivative of $0.4t+a$ as it's in the range of x=3.

finally, I have to find the expected value (E[x]) and I have no clue how to do that. would appreciate your help!

Answer 1

F(x)= 0 for t< 1 and $0.2t^2- 0.4t+ b$ for 1< t< 2. In order for the function to be continuous at t= 1, those must "match up": $F(1)= 0= 0.2(1^2)- 0.4(1)+ b= -0.2+ b$ so we must have $b= 0.2$. For 2< t< 4, F= 0.4t+ a. Again the two formulas must give the same value at t= 2. We must have $0=.2(2^2)- 0.4(2)+ 0.2= 0.4(2)+ a$. That reduces to $0.8- 0.8+ 0.2= 0.2= 0.8+ a$ so a= -0.6.

The expected value is $\int t f(t) dt$ where f is the probability density function, the derivative of the give $F_x(t)$.
That derivative is f(t)= 0 for t< 1, $0.4t- 0.4$ for $1\le t< 2$, $0.4$ for $2\le t< 4$ and 0 for $4\le t$.

So the expected value is given by $0.4\int_1^1t(t- 1)dt+ 0.4\int_2^4 t dt$. That should be easy to calculate.

Answer 2

Firstly, to have a continuous function $F$, you need to solve these equations which should be satisfied on the boundaries

$F(1)=0.2(1)^2-0.4(1)+b=0$

$F(2)=0.4(2)+a=0.2(2)^2-0.4(2)+b$

Then, either you can find the corresponding PDF or use another formula to find the expected value, which works for positive random variables

$\int\limits_0^{\infty} (1-F(t))dt$

Answer 3

We know for a continuous random variable, the PDF is the first derivative of the CDF wherever the CDF is differentiable. Here, as you see, the random variable is defined over the interval $ 1\le\ X \le\ 4 $ and the probability that it lies outside this interval is zero.
Now, within this interval, the CDF is a piecewise function with only one point where continuity/differentiability may break - namely the common endpoint of the subdomains within this interval, i.e. at $t=2$. So let us test for continuity and differentiability at this point. If we do this applying the usual tests of differentiability, we find the CDF is differentiable (which means also continuous) at $t=2$. Therefore the CDF is continuous throughout the interval over which the RV $X$ is distributed i.e. over $ 1\le\ X \le\ 4 $.
Now you can just differentiate the CDF to get the pdf for each interval. The PDF turns out to be:

$f(t) = 0.4(t-1), for$ $1\le\ t \lt\ 2$
$ $ $=0.4$ $for$ $ 2\le\ t \lt\ 4$

Now you can easily find $E(X)$ using:

$E(X)=\int_{1}^{4}xf(x)dx$

$=\int_{1}^{2}xf(x)dx + \int_{2}^{4}xf(x)dx$

$=\int_{1}^{2}0.4x(x-1)dx + 0.4\int_{2}^{4}dx$

$=0.4(\frac{8-1}{3}+\frac{4-1}{2}+2)$, if you do the calculations,

$=1.1333$