Computing probability of one random variable less than another — how to approach?
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I have 2 random variables X~U[-2,2], Y~U[0,1]. Lets define two new random variables $A=(X+Y)^2$, $B=X^2-Y^2$ .
What is the exact numerical value of the probability of $Pr(A<B)$?
My approach to solve this problem is to make a new random variable $Z=A-B$ and then to compute $F_Z(z)=Pr(Z<z)$ and finally solve for $F_z(0)=Pr(Z<0)=(A-B<0)$.
This is the only approach I can think of to solve this. Does this seem like an efficient approach?
probability-theory random-variables
asked Dec 25 '18 at 20:39
AvedisAvedis
647
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add a comment |
$begingroup$
I have 2 random variables X~U[-2,2], Y~U[0,1]. Lets define two new random variables $A=(X+Y)^2$, $B=X^2-Y^2$ .
What is the exact numerical value of the probability of $Pr(A<B)$?
My approach to solve this problem is to make a new random variable $Z=A-B$ and then to compute $F_Z(z)=Pr(Z<z)$ and finally solve for $F_z(0)=Pr(Z<0)=(A-B<0)$.
This is the only approach I can think of to solve this. Does this seem like an efficient approach?
probability-theory random-variables
asked Dec 25 '18 at 20:39
AvedisAvedis
647
$endgroup$
1
$begingroup$
Hint: $(X+Y)^2 - (X^2 - Y^2) = X^2 + 2XY + Y^2 - X^2 + Y^2 = 2(XY + 2Y^2)$. This is negative exactly when $X+Y$ is negative (because $2Y$ is nonnegative). Now you can just draw a picture of the rectangle $[-2, 2] times [0, 1]$ and shade in various parts of it.
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– John Hughes
Dec 25 '18 at 20:42
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Thanks. I actually factored it to be P(X+Y<0), graphed, and then shaded in the region of the left of the line which intersects the pdfs of X and Y. The pdf have am amplitude of 4. Integrating to the left of P(X+Y<0) you get (1+1/2)/4 = 3/8
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– Avedis
Dec 25 '18 at 21:01
add a comment |
$begingroup$
I have 2 random variables X~U[-2,2], Y~U[0,1]. Lets define two new random variables $A=(X+Y)^2$, $B=X^2-Y^2$ .
What is the exact numerical value of the probability of $Pr(A<B)$?
My approach to solve this problem is to make a new random variable $Z=A-B$ and then to compute $F_Z(z)=Pr(Z<z)$ and finally solve for $F_z(0)=Pr(Z<0)=(A-B<0)$.
This is the only approach I can think of to solve this. Does this seem like an efficient approach?
probability-theory random-variables
asked Dec 25 '18 at 20:39
AvedisAvedis
647
$endgroup$
I have 2 random variables X~U[-2,2], Y~U[0,1]. Lets define two new random variables $A=(X+Y)^2$, $B=X^2-Y^2$ .
What is the exact numerical value of the probability of $Pr(A<B)$?
My approach to solve this problem is to make a new random variable $Z=A-B$ and then to compute $F_Z(z)=Pr(Z<z)$ and finally solve for $F_z(0)=Pr(Z<0)=(A-B<0)$.
This is the only approach I can think of to solve this. Does this seem like an efficient approach?
probability-theory random-variables
probability-theory random-variables
asked Dec 25 '18 at 20:39
AvedisAvedis
647
asked Dec 25 '18 at 20:39
AvedisAvedis
647
asked Dec 25 '18 at 20:39
AvedisAvedis
647
asked Dec 25 '18 at 20:39
AvedisAvedis
647
asked Dec 25 '18 at 20:39
AvedisAvedis
647
647
1
$begingroup$
Hint: $(X+Y)^2 - (X^2 - Y^2) = X^2 + 2XY + Y^2 - X^2 + Y^2 = 2(XY + 2Y^2)$. This is negative exactly when $X+Y$ is negative (because $2Y$ is nonnegative). Now you can just draw a picture of the rectangle $[-2, 2] times [0, 1]$ and shade in various parts of it.
$endgroup$
– John Hughes
Dec 25 '18 at 20:42
$begingroup$
Thanks. I actually factored it to be P(X+Y<0), graphed, and then shaded in the region of the left of the line which intersects the pdfs of X and Y. The pdf have am amplitude of 4. Integrating to the left of P(X+Y<0) you get (1+1/2)/4 = 3/8
$endgroup$
– Avedis
Dec 25 '18 at 21:01
add a comment |
1
$begingroup$
Hint: $(X+Y)^2 - (X^2 - Y^2) = X^2 + 2XY + Y^2 - X^2 + Y^2 = 2(XY + 2Y^2)$. This is negative exactly when $X+Y$ is negative (because $2Y$ is nonnegative). Now you can just draw a picture of the rectangle $[-2, 2] times [0, 1]$ and shade in various parts of it.
$endgroup$
– John Hughes
Dec 25 '18 at 20:42
$begingroup$
Thanks. I actually factored it to be P(X+Y<0), graphed, and then shaded in the region of the left of the line which intersects the pdfs of X and Y. The pdf have am amplitude of 4. Integrating to the left of P(X+Y<0) you get (1+1/2)/4 = 3/8
$endgroup$
– Avedis
Dec 25 '18 at 21:01
1
1
$begingroup$
Hint: $(X+Y)^2 - (X^2 - Y^2) = X^2 + 2XY + Y^2 - X^2 + Y^2 = 2(XY + 2Y^2)$. This is negative exactly when $X+Y$ is negative (because $2Y$ is nonnegative). Now you can just draw a picture of the rectangle $[-2, 2] times [0, 1]$ and shade in various parts of it.
$endgroup$
– John Hughes
Dec 25 '18 at 20:42
$begingroup$
Hint: $(X+Y)^2 - (X^2 - Y^2) = X^2 + 2XY + Y^2 - X^2 + Y^2 = 2(XY + 2Y^2)$. This is negative exactly when $X+Y$ is negative (because $2Y$ is nonnegative). Now you can just draw a picture of the rectangle $[-2, 2] times [0, 1]$ and shade in various parts of it.
$endgroup$
– John Hughes
Dec 25 '18 at 20:42
$begingroup$
Thanks. I actually factored it to be P(X+Y<0), graphed, and then shaded in the region of the left of the line which intersects the pdfs of X and Y. The pdf have am amplitude of 4. Integrating to the left of P(X+Y<0) you get (1+1/2)/4 = 3/8
$endgroup$
– Avedis
Dec 25 '18 at 21:01
$begingroup$
Thanks. I actually factored it to be P(X+Y<0), graphed, and then shaded in the region of the left of the line which intersects the pdfs of X and Y. The pdf have am amplitude of 4. Integrating to the left of P(X+Y<0) you get (1+1/2)/4 = 3/8
$endgroup$
– Avedis
Dec 25 '18 at 21:01
add a comment |
1 Answer
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Yes it is. We have $$Pr(Z<0){=Pr(A<B)\=Pr(2XY+Y^2<-Y^2)\=Pr(X<-Y)\={3over 8}}$$
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
$endgroup$
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
add a comment |
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1 Answer
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Yes it is. We have $$Pr(Z<0){=Pr(A<B)\=Pr(2XY+Y^2<-Y^2)\=Pr(X<-Y)\={3over 8}}$$
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
$endgroup$
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
add a comment |
$begingroup$
Yes it is. We have $$Pr(Z<0){=Pr(A<B)\=Pr(2XY+Y^2<-Y^2)\=Pr(X<-Y)\={3over 8}}$$
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
$endgroup$
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
add a comment |
$begingroup$
Yes it is. We have $$Pr(Z<0){=Pr(A<B)\=Pr(2XY+Y^2<-Y^2)\=Pr(X<-Y)\={3over 8}}$$
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
$endgroup$
Yes it is. We have $$Pr(Z<0){=Pr(A<B)\=Pr(2XY+Y^2<-Y^2)\=Pr(X<-Y)\={3over 8}}$$
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
answered Dec 25 '18 at 20:45
Mostafa AyazMostafa Ayaz
15.7k3939
15.7k3939
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
add a comment |
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
$begingroup$
I just computed this on my own and got 3/8. Thanks!
$endgroup$
– Avedis
Dec 25 '18 at 20:56
add a comment |
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$begingroup$
Hint: $(X+Y)^2 - (X^2 - Y^2) = X^2 + 2XY + Y^2 - X^2 + Y^2 = 2(XY + 2Y^2)$. This is negative exactly when $X+Y$ is negative (because $2Y$ is nonnegative). Now you can just draw a picture of the rectangle $[-2, 2] times [0, 1]$ and shade in various parts of it.
$endgroup$
– John Hughes
Dec 25 '18 at 20:42
$begingroup$
Thanks. I actually factored it to be P(X+Y<0), graphed, and then shaded in the region of the left of the line which intersects the pdfs of X and Y. The pdf have am amplitude of 4. Integrating to the left of P(X+Y<0) you get (1+1/2)/4 = 3/8
$endgroup$
– Avedis
Dec 25 '18 at 21:01