Probability of a rectangle having a perimeter of 5

[jeffo]

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.

 

[blamocur]

I would compute the 2D density function [imath]g(x,y)[/imath].
Then figure out the set [imath]S[/imath] of all [imath](x,y)[/imath] for which the perimeter of the corresponding rectangle is less than 5.
Using the density function you can find the probability of [imath]x,y[/imath] being inside set [imath]S[/imath].

Does this help?

 

[BigBeachBanana]

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
[math]\Pr(2X+2Y<5)=\Pr(X+Y<5/2)=\iint_{R}f(x,y)\,dR[/math]Since X and Y are independent, [imath]f(x,y)=f(x)f(y)[/imath]
I'll let you figure the region R.
Hint:
1)Graph [imath]y=5/2-x[/imath]
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?

 

[jeffo]

The goal is to find
[math]\Pr(2X+2Y<5)=\Pr(X+Y<5/2)=\iint_{R}f(x,y)\,dR[/math]Since X and Y are independent, [imath]f(x,y)=f(x)f(y)[/imath]
I'll let you figure the region R.
Hint:
1)Graph [imath]y=5/2-x[/imath]
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

 

[jeffo]

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

 

[BigBeachBanana]

5/2 - y* for x limits

Correct. The region R is bounded by y=0,x=0 and x+y=5/2

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.
[math]\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)[/math] In other words, it's not negligible, but it's precisely 0.