Pairs of Random Variables Question

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rsingh628

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Hello, I would like to check my understanding and get some help with last part of the following question, please.
For part (d), would I use f(x | y) = f(x, y) / f(y) ? Thank you for your time.

1662306451784.png

My attempt so far is shown below:

1662307952244.png
 
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rsingh628 said:
Hello, I would like to check my understanding and get some help with last part of the following question, please.
For part (d), would I use f(x | y) = f(x, y) / f(y) ? Thank you for your time.

View attachment 33942

My attempt so far is shown below:

View attachment 33944
Click to expand...
For (d), draw the integration region [imath]R[/imath] i.e. [imath]X+Y \le 1[/imath] and [imath]Y\le 0.5[/imath] to determine your limits of integration.

[math]\Pr(X+Y\le 1 | Y\le 0.5)= \iint_R f(x,y)\, dR[/math]
 
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BigBeachBanana said:
For (d), draw the integration region [imath]R[/imath] i.e. [imath]X+Y \le 1[/imath] and [imath]Y\le 0.5[/imath] to determine your limits of integration.

[math]\Pr(X+Y\le 1 | Y\le 0.5)= \iint_R f(x,y)\, dR[/math]
Click to expand...

Am I on the right track with this set-up, or am I totally off? I would need the marginal pdf for Y for the denominator.

1662322623305.png
 
rsingh628 said:
Am I on the right track with this set-up, or am I totally off? I would need the marginal pdf for Y for the denominator.

View attachment 33946
Click to expand...
I haven't checked your integrals yet, but if you draw the integrating region you don't need to use the conditional density because the bounds took care of the conditional part.

Screen Shot 2022-09-04 at 5.10.04 PM.png
 
rsingh628 said:
Hello, I would like to check my understanding and get some help with last part of the following question, please.
For part (d), would I use f(x | y) = f(x, y) / f(y) ? Thank you for your time.


My attempt so far is shown below:

View attachment 33944
Click to expand...
In the 2nd line in part a, you have an error in what you wrote after the 1st equal sign.


2nd from last line:
x2(1-x)3
In (1-x)3 after expanding one term will be 1. When you multiply x2 by 1 you get x2. However you got x^3 -2x^4 + x^5. This result is totally wrong.
 
Steven G said:
In the 2nd line in part a, you have an error in what you wrote after the 1st equal sign.


2nd from last line:
x2(1-x)3
In (1-x)3 after expanding one term will be 1. When you multiply x2 by 1 you get x2. However you got x^3 -2x^4 + x^5. This result is totally wrong.
Click to expand...
I am guessing that [imath]x^2(1-x)^3[/imath] was replaced by [imath]x^3(1-x)^2[/imath], which would have the same integral on the [0,1] interval.
 
rsingh628 said:
Am I on the right track with this set-up, or am I totally off? I would need the marginal pdf for Y for the denominator.

View attachment 33946
Click to expand...
Part d:
You have \(\displaystyle \int_{0.5}^1 [\int_{0.5}^1 24x^2(1-x)dx]dy\)
That inner integral will be a constant, which can be put in front of the remaining integral. This constant will be exactly what we are dividing by so there is not need to even evaluate this integral (except if it equals 0 which it clearly doesn't) as it will cancel out to 1 with the denominator.

Edit: The inner integral does not have to even be a constant to cancel out with the denominator---it just has to be NOT a function of y.
 
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Steven G said:
In the 2nd line in part a, you have an error in what you wrote after the 1st equal sign.


2nd from last line:
x2(1-x)3
In (1-x)3 after expanding one term will be 1. When you multiply x2 by 1 you get x2. However you got x^3 -2x^4 + x^5. This result is totally wrong.
Click to expand...
My mistake here. I revised my work for part (c). The answer is still 1/5 or 0.2, I believe. would you agree?

1662388146554.png
 
Steven G said:
Part d:
You have \(\displaystyle \int_{0.5}^1 [\int_{0.5}^1 24x^2(1-x)dx]dy\)
That inner integral will be a constant, which can be put in front of the remaining integral. This constant will be exactly what we are dividing by so there is not need to even evaluate this integral (except if it equals 0 which it clearly doesn't) as it will cancel out to 1 with the denominator.

Edit: The inner integral does not have to even be a constant to cancel out with the denominator---it just has to be NOT a function of y.
Click to expand...

Okay, so I feel confident with my answers for part a - c.
Now, for part (d), my attempt is as follows. Do we agree here? Thanks for your help by the way.

1662501421274.png
 
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