Probability of a rectangle having a perimeter of 5

jeffo

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Hi,
If I was given a rectangle with the probability density functions of the width being f(x) = e^(-x) and height of f(y) = 0.5e^(-0.5y), how would I find the probability of the perimeter of the rectangle being less than 5m?
Do I combine this into 1 perimeter function (P(x,y) = 2(e^(-x)) + 2(0.5e^(-0.5y)) ) ?
Appreciate any help.
 
I would compute the 2D density function g(x,y)g(x,y).
Then figure out the set SS of all (x,y)(x,y) for which the perimeter of the corresponding rectangle is less than 5.
Using the density function you can find the probability of x,yx,y being inside set SS.

Does this help?
 
Hi,
If I was given a rectangle with the probability density functions of the width being f(x) = e^(-x) and height of f(y) = 0.5e^(-0.5y), how would I find the probability of the perimeter of the rectangle being less than 5m?
Do I combine this into 1 perimeter function (P(x,y) = 2(e^(-x)) + 2(0.5e^(-0.5y)) ) ?
Appreciate any help.
The goal is to find
Pr(2X+2Y<5)=Pr(X+Y<5/2)=Rf(x,y)dR\Pr(2X+2Y<5)=\Pr(X+Y<5/2)=\iint_{R}f(x,y)\,dRSince X and Y are independent, f(x,y)=f(x)f(y)f(x,y)=f(x)f(y)
I'll let you figure the region R.
Hint:
1)Graph y=5/2xy=5/2-x
2) Notice f(x) and f(y) are exponential distributions, consider their domains. Or from a practical standpoint, for what values of x and y does not make sense?
 
Last edited:
The goal is to find
Pr(2X+2Y<5)=Pr(X+Y<5/2)=Rf(x,y)dR\Pr(2X+2Y<5)=\Pr(X+Y<5/2)=\iint_{R}f(x,y)\,dRSince X and Y are independent, f(x,y)=f(x)f(y)f(x,y)=f(x)f(y)
I'll let you figure the region R.
Hint:
1)Graph y=5/2xy=5/2-x
2) Notice f(x) and f(y) are exponential distributions, consider their domains. Or from a practical standpoint, for what values of x and y does not make sense?
Thank you for the help!
So the region I would integrate over would be for 0 to 5/2-x for x limits and 0 to 5/2 for y limits, R=dxdy?
If I am not mistaken, since the limits are inclusive, would this give Pr(2X+2Y<=5)?
Is it ok to assume that the probability that perimeter = 5 is negligible?
Thank you again
 
Thank you for the help!
So the region I would integrate over would be for 0 to 5/2-x for x limits and 0 to 5/2 for y limits, R=dxdy?
If I am not mistaken, since the limits are inclusive, would this give Pr(2X+2Y<=5)?
Is it ok to assume that the probability that perimeter = 5 is negligible?
Thank you again
5/2 - y* for x limits
 
5/2 - y* for x limits
Correct. The region R is bounded by y=0,x=0 and x+y=5/2
1649268059040.png
You have two options: Integrate over x, then y OR y then x.

In the OP, you wrote: "find the probability of the perimeter of the rectangle [is] less than 5m, not less than or equal to 5.
In either case, it doesn't impact the result because, in general, for continuous random variable W.
Pr(Ww)=Pr(W=w)+Pr(W<w)=wwf(w)dw+wf(w)dw=0+Pr(W<w)    Pr(Ww)=Pr(W<w)\Pr(W\le w)=\Pr(W = w)+\Pr(W < w)= \int_{w}^{w}f(w)\,dw+\int_{-\infty}^{w}f(w)\,dw=0+\Pr(W < w)\\ \implies \Pr(W\le w)=\Pr(W < w) In other words, it's not negligible, but it's precisely 0.
 
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