Integration of the expected value

Dear community!

I have an interval and an expected price which is laying on this interval. b is a constant.

I want to find all possible values of expected profit function which is expressed as and plot it.

is a c.d.f. of uniformly distrtubited density function .

Integrating by parts the expected price can be re-written as: . So, I am trying to write the following code, which doesn't work:, please see below. Any help will be highly apprerciated!

b = 1;
p = (0:0.01:b);
p_e = (p:0.01:b);
pd1 = makedist('Uniform','lower',p,'upper',b);

G = @(p) p + p * cdf - int((cdf), p, p, b);
y=G(p);

plot(p,y)
xlim([0 1]) 
ylim([0 +inf])

leg = legend('Expected profit','AutoUpdate','off');
title(leg,'Expected profit')
xlabel('Price')
ylabel('Expected Profit')

% put the grid on top of the colored area
set(gca, 'Layer', 'top')
grid on

Answer 1

[Torsten]

b = 1;
N = 2000000;
for i = 1:N
    p(i) = b*rand;
    pe(i) = p(i) + (b-p(i))*rand;
    p2(i) = p(i) + pe(i);
end
ecdf(pe)

mean(p)

ans = 0.5000

mean(pe)

ans = 0.7499

mean(p2)

ans = 1.2499

[Torsten]

The expectation for pe, E[pe], is easy to compute:

pe = b*X + (b-b*X)*Y

where X and Y are independent random variables uniformly distributed on [0;1].

Thus

E[pe] = b*E[X] + b*E[Y] - b*E[X]*E[Y] = b*0.5 + b*0.5 - b*0.5*0.5 = 0.75*b

If someone could compute the pdf or cdf - Hats off !

[Yuriy]

Hi! Thank you very much! Could you please explain me what means N = 2000000?

does it mean that you draw random value from 1 to 2000000? Because b in my question is an upper bound of the support, and from my code you can see that I was trying to set up an interval from 0 to 1 with a step 0.01.

[Torsten]

The code is a Monte-Carlo simulation to get the cdf of the random variable pe.

For this purpose, I used many (2 x 2000000 in this case) draws from a uniform distribution. With the first draw, I determined p (uniformly distributed between 0 and b). With the second draw, I determined pe uniformly distributed between p and b. Then I let MATLAB calculate the empirical cdf of pe.

As I wrote, the expected value for pe can easily be deduced by the above calculation if the complete cdf for pe does not matter.

[Yuriy]

I see, thanks! I still wonder how to plot all a profit curve... (if it's possible)

[Torsten]

If pi = p+pe is the profit, you can plot a probability distribution (pdf) for pi. It would look like

b = 1;
N = 20000000;
for i = 1:N
    p(i) = b*rand;
    pe(i) = p(i) + (b-p(i))*rand;
    p2(i) = p(i) + pe(i);
end
ksdensity(p2)
xlim([0,2])

[Yuriy]

Great, thanks! Not sure if this answers my question, but I mark it because looks like it is the closest. Thank you! I think I need to reformulate my question. I need to draw 2 random variables p1 and p2 such that p1 is not equal to p2. And then my expecteds profit will be (p1 + p2) or (p1 + b) depends on which is greater (dominating). I will try to plot it too right now, but any help will be much appreciated becasue I am new to this and my learning curve is not the greatest, unfortunately.

Answer 2

[William Rose]

@Yuriy,

I'm not sure I understand. Is a uniformly distributed random variable, or is the expected value of p, which is a uniformly distributed random variable? You wrote inside the integral in the first equation, but after that, the subscript e disappeared. If is the random variable inside the integral, then it should be the vrable of integration, so the integral should have been . If that is the case, then you have

where b and p are constants. Which is just what you expect for the mean value of a uniformly distributed random variable.

Perhaps you meant to write for the first equation. If so, there is a different problem, which is that p cannot be the lower limit of integration and the variable of integration at the same time.

[Yuriy]

I apologize for my messy explanation. A variable in the first period denoted , and in the second is . Therefore, .

So, in the first period I randomly draw p from the distribution with the support , and then in the second period I anticapate price raise, which will be from the same distribution but with the support . Therefore, I am trying to plot expected profits curve.

I see what you mean, and it is what MatLab saying about this piece of code:

p_e = (p:0.01:b);

But the expected price will depend on how price p changes the interval over which I am trying to intergate.