# Parametric Region Plot

I tried to find solution in a few discussions but I couldn't solve my problem because my knowledge is basic.

I have the following functions and parameters:

Mstar[d_, u_, p_] := ProductLog[-d p + p u];

PIstar[d_, u_, p_] := u - E^-Mstar[d, u, p] (u - d);

CMstar[d_, u_, p_] := 1/2 Mstar[d, u, p]^2;

EUY1star[d_, u_, p_] := PIstar[d, u, p] p - CMstar[d, u, p];

EUB1star[d_, u_, p_, r_, s_] := r (1 - PIstar[d, u, p]) - 2 PIstar[d, u, p] s r;

EUB2star[d_, u_, p_, r_, s_] := r (1 - d) - 3 d s r - EUY1star[d, u, p];

d = 0.2;
u = 0.8;
r = 1;


and I want to draw the Parametric Region Plot of

[{EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] > 0 && EUB2star[d, u, p, r, s] > 0}, {p, 0, 1000}]

where in the horizontal axis I need PIstar[d, u, p] and in the vertical axis {s, 0, 2}.

I have no idea how to do it yet in my V.8. Pls help. Maurizio.

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• For the given value of {d,u,r}, the condition {EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] > 0 && EUB2star[d, u, p, r, s] > 0} is not True for any (p,s) in the range you have given. Also PIstar[d, u, p] is complex. – Hubble07 Sep 13 '15 at 12:20
• Hi Hubble07. I am a bit puzzled about your answer. I tell you why. Originally I plotted the following non-overlapping regions: RegionPlot[{EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] > 0 && EUB2star[d, u, p, r, s] > 0, EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] < 0 && EUB1star[d, u, p, r, s] > 0, EUB1star[d, u, p, r, s] < 0 && EUB2star[d, u, p, r, s] < 0}, {p, 0, 10}, {s, 0, 2}, PlotStyle -> {Red, Blue, Green}, MaxRecursion -> 10] – M.Lisciandra Sep 13 '15 at 13:00
• @M.Lisciandra Sorry, I made a silly mistake by checking only integer values using Table. Anyway checkout the answer i posted. – Hubble07 Sep 13 '15 at 14:55

RegionPlot[{EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] > 0 &&
EUB2star[d, u, p, r, s] > 0},
{p, 0, 2}, {s, 0, 2}]


• The OP said they need PIstar[d, u, p] along the X-axis not simply p. – Hubble07 Sep 13 '15 at 14:59
• Yes. Indeed. Thank you Belisarius by the way. – M.Lisciandra Sep 13 '15 at 15:08

## Version 2

How about writing all your function in terms of $\pi^*$ and then plotting them from 0.2 to 0.8 like this

 region = RegionPlot[{r (1 - d) -
3 d s r - (pstar*(Log[(u - d)/(u - pstar)]/(u -
pstar)) - (Log[(u - d)/(u - pstar)])) - (r (1 - pstar) -
2 pstar*s r) > 0 &&
r (1 - d) -
3 d s r - (pstar (Log[(u - d)/(u - pstar)]/(u -
pstar)) - (Log[(u - d)/(u - pstar)])) > 0}, {pstar, 0.2,
0.8}, {s, 0, 2}, MaxRecursion -> 10,
FrameLabel -> {"\!$$\*SuperscriptBox[\(\[Pi]$$, $$*$$]\)", "s"},
RotateLabel -> False, LabelStyle -> (FontSize -> 20)]


For the transformation I used the inverse function for ProductLog

If $b = ProductLog(a)$ then $a = b*e^{b}$

Using above we can write

$$p(u-d)=M^* e^{M^*}$$

This allows to write $p$ in terms of $\pi^*$

$$M^* = Log\left(\frac{u-d}{u-\pi^*}\right)$$

$$p= \frac{1}{u-\pi^*}Log\left(\frac{u-d}{u-\pi^*}\right)$$

With the above transformation you can express all your inequality in terms of $\pi^*$. Since $\pi^*$ ranges from $u$ to $d$ for $p$ going from $0$ to $\infty$, you can then RegionPlot $\pi^*$ from u to d. This is what I have shown.

Can't you just find the range of $\pi^*$ for the range of $p$ and then plot over that range.

  regPlot[pmax_, smax_] :=
Module[{pmin = 0, smin = 0, PIstarMin, PIstarMax},

PIstarMin = PIstar[d, u, pmin];
PIstarMax = PIstar[d, u, pmax];

RegionPlot[{EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] > 0 &&
EUB2star[d, u, p, r, s] > 0}, {p, PIstarMin, PIstarMax}, {s, 0,
smax}, MaxRecursion -> 10,
FrameLabel -> {"\!$$\*SuperscriptBox[\(\[Pi]$$, $$*$$]\)", "s"},
RotateLabel -> False, LabelStyle -> (FontSize -> 20)]]


So for $p_{max}=1000$ and $s_{max}=2$ we get

   regPlot[1000, 2]


• Dear Hubble07, you're just great! I have a question though. The variable 'p' goes from 0 to infinity (because this is my definition of the variable). I put p=[0,1000] just for simplicity. 'PIstar' is a probability and it is simply a function of 'p', given 'd' and 'u', which are the lower and upper bounds of 'PIstar' respectively. Thus 'PIstar' is fully described by the whole range of 'p=[0,+infinity or 1000]'. Hence, the plot you gave me describes 'PIstar' values for 'p' that goes from 0 to 1000. Is it correct? Thank you again!! – M.Lisciandra Sep 13 '15 at 16:02
• Thanks. Actually in the plot $\pi^*$ goes upto 0.8 which is infact the limiting value for $p \rightarrow \infty$. I mean $lim_{p \rightarrow \infty}\pi^*(a,b,p)=b$. You can check this in mma Limit[PIstar[a, b, p], p -> Infinity]. Here its 0.8 since you have set that limit by u=0.8. So you can sat that, "The plot describes $\pi^*$ values for $p$ going from 0 to $\infty$" – Hubble07 Sep 13 '15 at 16:25
• I'm afraid there is a problem with the solution you gave me! I did a double check. First, consider simply EUB1star>0, I plotted parametrically the boundary ParametricPlot[{PIstar[d, u, p], (1 - PIstar[d, u, p])/(2 PIstar[d, u, p])}, {p, 0, 1000}, AspectRatio -> 1/1, PlotRange -> {{d, u}, {0, 2}}, MaxRecursion -> 15, AxesOrigin -> {d, 0}] ... the region plot is actually the area beneath the curve. Then, I used the regionplot you suggested just for EUB1star[d, u, p, r, s] > 0 but the areas are different! Any suggestions? Sorry if I take some of your time.... Thank you. – M.Lisciandra Sep 13 '15 at 21:32
• I've just realized that with your script you've just plotted simply this: RegionPlot [{EUB2star[d, u, p, r, s] - EUB1star[d, u, p, r, s] > 0 && EUB2star[d, u, p, r, s] > 0}, {p, 0.2, 0.8}, {s, 0, 2}, MaxRecursion -> 10], as @belisarius did before. – M.Lisciandra Sep 13 '15 at 21:44
• I don't see any problem in Regionplot for EUB1star[d, u, p, r, s] >0 given by my function. What do you expect?. It agrees with Plot[EUB1star[d, u, 0.2, r, s], {s, 0, 2}]. I mean the maximum s value allowed is around 1.4. In your ParametricPlot whats the expected value of Y-coordinate at p=0.2. It should be the value of (1 - PIstar[d, u, 0.2])/(2 PIstar[d, u, 0.2]) right which is around 1.4 so it matches with the cutoff in the RegionPlot. – Hubble07 Sep 13 '15 at 23:29