Finding P(X1<X2<X3) and P(X1=X2<X3)

[cookiesyum]

Homework Statement

Let f(x₁, x₂, x₃) = e^{-(x₁+x₂+x₃)}, 0<x_1,2,3<infinity, zero elsewhere be a joint pdf of X₁, X₂, X₃.

Compute P(X₁< X₂< X₃) and P(X₁= X₂< X₃)

Determine the joint mgf.

The Attempt at a Solution

P(X₁< X₂< X₃) = triple integral of e^{-(x₁+x₂+x₃)} dx₂dx₁dx₃ as x₂ goes from x₁ to x₃, x₁ goes from 0 to x₂ and x₃ goes from x₂ to infinity.

When I solve this integral I get 0 for an answer. The back of the book suggests the answer is 1/6. Clearly, I am setting up the problem wrong, but I don't know where my mistake is. Similarly, I'm having trouble setting up the second part of the question as well.

Joint mgf = E(e^{t1x₁+t2x₂+t3x₃}) but this integral is not easy by hand. Is there another way to do it?

 

[clamtrox]

Yeah, your integration limits are incorrect. For P(x_1<x_2<x_3), start from x_1. Say it goes from zero to infinity. Then x_2 has to go from x_1 to infinity, and likewise x_3 goes from x_2 to infinity.

What's hard about solving the MGF?

$$E(e^{t(x_1+x_2+x_3)}) = \int dx_1 dx_2 dx_3 f(x_1,x_2,x_3) e^{t(x_1+x_2+x_3)} = \int dx_1 dx_2 dx_3 \exp [(t-1)(x_1+x_2+x_3)]$$

I'm sure you know how to integrate an exponential function.

 

[cookiesyum]

Yeah, your integration limits are incorrect. For P(x_1<x_2<x_3), start from x_1. Say it goes from zero to infinity. Then x_2 has to go from x_1 to infinity, and likewise x_3 goes from x_2 to infinity.

What's hard about solving the MGF?

$$E(e^{t(x_1+x_2+x_3)}) = \int dx_1 dx_2 dx_3 f(x_1,x_2,x_3) e^{t(x_1+x_2+x_3)} = \int dx_1 dx_2 dx_3 \exp [(t-1)(x_1+x_2+x_3)]$$

I'm sure you know how to integrate an exponential function.

Oh! I was doing it with three different t's ... t1 t2 and t3 which made it a million times harder. That makes sense. Thanks so much!

 

[clamtrox]

I guess you do need multiple t's. It doesn't make it any harder though.

 

[statdad]

No, you don't need multiple ts.

$$\phi(t)_{X_1+X_2+X_3} = \idotsint e^{t(x_1+x_2+x_3)} f(x_1,x_2,x_3)\,dx_1dx_2dx_3$$

Aren't your variables independent exponentials?

 

[cookiesyum]

No, you don't need multiple ts.

$$\phi(t)_{X_1+X_2+X_3} = \idotsint e^{t(x_1+x_2+x_3)} f(x_1,x_2,x_3)\,dx_1dx_2dx_3$$

Aren't your variables independent exponentials?

Yep, finally got the answer. Thank you everybody for your help!

 

[mbkemp23]

What about P(X1=X2<X3)?

 

[Mute]

What about P(X1=X2<X3)?

The random variables are drawn from a continuous distribution, and are independent. Do you know what the probability is that a continuous random variable will be exactly some specified number?

 

[mbkemp23]

zero, it must be an interval... So then would it be P(X1=X2<X3) = P(X1<X3)*P(X2<X3) or even P(0<X3)?

 

[Ray Vickson]

Homework Statement

Let f(x₁, x₂, x₃) = e^{-(x₁+x₂+x₃)}, 0<x_1,2,3<infinity, zero elsewhere be a joint pdf of X₁, X₂, X₃.

Compute P(X₁< X₂< X₃) and P(X₁= X₂< X₃)

Determine the joint mgf.

The Attempt at a Solution

P(X₁< X₂< X₃) = triple integral of e^{-(x₁+x₂+x₃)} dx₂dx₁dx₃ as x₂ goes from x₁ to x₃, x₁ goes from 0 to x₂ and x₃ goes from x₂ to infinity.

When I solve this integral I get 0 for an answer. The back of the book suggests the answer is 1/6. Clearly, I am setting up the problem wrong, but I don't know where my mistake is. Similarly, I'm having trouble setting up the second part of the question as well.

Joint mgf = E(e^{t1x₁+t2x₂+t3x₃}) but this integral is not easy by hand. Is there another way to do it?

You are correct that this is the joint mgf (with three different t's). If you had t1 = t2 = t3 = t, as some others have suggested, you would be finding the mgf of the single random variable X = X1 + X2 + X3. Anyway, the integral is *very easy* by hand.

RGV

 

[Artusartos]

You are correct that this is the joint mgf (with three different t's). If you had t1 = t2 = t3 = t, as some others have suggested, you would be finding the mgf of the single random variable X = X1 + X2 + X3. Anyway, the integral is *very easy* by hand.

RGV

I have a question about the mgf...

So this is the integral that we need to integrate, right?

$\int_0^{\infty} \int_0^{\infty} \int_0^{\infty} e^{x_1(t_1-1)} e^{x_2(t_2-1)} e^{x_3(t_3-1)} dx_1 dx_2 dx_3$

But I'm having some trouble with this, because this is what I get after integrating...

$\frac{1}{t_1-1}e^{x_1(t_1-1)} |_0^{\infty} \frac{1}{t_2-1}e^{x_2(t_2-1)} |_0^{\infty} \frac{1}{t_3-1}e^{x_3(t_3-1)} |_0^{\infty}$...but if we plug in infinity to $\frac{1}{t_1-1}e^{\infty(t_1-1)}$...don't we get infinity, since e increases forever?

 

[clamtrox]

I have a question about the mgf...

So this is the integral that we need to integrate, right?

$\int_0^{\infty} \int_0^{\infty} \int_0^{\infty} e^{x_1(t_1-1)} e^{x_2(t_2-1)} e^{x_3(t_3-1)} dx_1 dx_2 dx_3$

But I'm having some trouble with this, because this is what I get after integrating...

$\frac{1}{t_1-1}e^{x_1(t_1-1)} |_0^{\infty} \frac{1}{t_2-1}e^{x_2(t_2-1)} |_0^{\infty} \frac{1}{t_3-1}e^{x_3(t_3-1)} |_0^{\infty}$...but if we plug in infinity to $\frac{1}{t_1-1}e^{\infty(t_1-1)}$...don't we get infinity, since e increases forever?

That depends on whether t-1 is positive or negative.

 

[Artusartos]

That depends on whether t-1 is positive or negative.

Thanks a lot...