Problem 15 Let $X_{1}, X_{2}$ be two rand... [FREE SOLUTION]

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Introduction to Mathematical Statistics

Robert V. Hogg, Joeseph McKean, Allen T Craig

$Math Studyset 91影视 Explanations$ Math

7 Edition

Chapter 2: Problem 15

Let $X_{1}, X_{2}$ be two random variables with joint $\operatorname{pdf} f\left(x_{1}, x_{2}\right)=x_{1} \exp \left\\{-x_{2}\right\\}$, for \(0

Short Answer

Expert verified

The joint moment generating function $M(t_{1}, t_{2})$ of $X_{1}, X_{2}$ is $\frac{1}{(1-t_{1})(2-t_{2})}$ for $t_{1}<1$ and $t_{2}<2$, and undefined elsewhere. The statement $M(t_{1}, t_{2}) = M(t_{1}, 0) M(0, t_{2})$ is incorrect.

Step by step solution

Setting up the joint mgf

The joint mgf $M(t_{1}, t_{2})$ is defined as $E[e^{t_{1}X_{1} + t_{2}X_{2}}]$. It can be computed as a double integral over the pdf, this results to: \[M(t_{1}, t_{2})= \int_0^\infty\int_{x_{1}}^\infty e^{t_{1}x_{1} + t_{2}x_{2}} f(x_{1},x_{2}) dx_{2} dx_{1}\]

Substituting the pdf

Now insert the given pdf $f(x_{1},x_{2})=x_{1} \exp \left\{-x_{2}\right\}$ into the integral, to get: \[M(t_{1}, t_{2})= \int_0^\infty\int_{x_{1}}^\infty e^{t_{1}x_{1} + t_{2}x_{2}} x_{1} e^{-x_{2}} dx_{2} dx_{1}\]

Simplifying the integral

This can be simplified using exponent properties, which turns into: \[M(t_{1}, t_{2})= \int_0^\infty\int_{x_{1}}^\infty x_{1} e^{(t_{1}-1)x_{1} + (t_{2}-1)x_{2}} dx_{2} dx_{1}\]

Solving the integral

We now solve the inner integral, which will give us: \[ M(t_{1}, t_{2})= \int_0^\infty x_{1} e^{(t_{1}-1)x_{1}} \left[\int_{x_{1}}^\infty e^{ (t_{2}-1)x_{2}} dx_{2} \right] dx_{1}\] The inner integral evaluates to \[-\frac{1}{t_{2}-1} e^{(t_{2}-1)x_{2}} \Biggr|_{x_{1}}^{\infty} = - \frac{e^{(t_{2}-1)x_{1}}}{t_{2}-1} \] Substituting this result back into the integral we have: \[M(t_{1}, t_{2}) = \int_0^\infty -\frac{x_{1}e^{(t_{1}+t_{2}-2)x_{1}}}{t_{2}-1} dx_{1}\]

Solving the remaining integral

The remaining integral can be solved in a similar manner, resulting to: \[M(t_{1}, t_{2}) = -\frac{e^{(t_{1}+t_{2}-2)x_{1}} }{(t_{1}+t_{2}-2)(t_{2}-1)} \Biggr|_0^\infty \] which gives $M(t_{1}, t_{2}) = \frac{1}{(1-t_{1})(2-t_{2})}$ for $t_{1}<1$ and $t_{2}<2$, and undefined elsewhere

Checking the equivalence

We now compare this with the right-hand side of the given equivalence(which is the product of the marginals) $M(t_{1}, 0)M(0, t_{2})$ to check if they are equal. It turns out they are not equal because there is no t_{2} in $M(t_{1}, 0)$ and no t_{1} in $M(0, t_{2})$, so $M(t_{1}, t_{2}) \neq M(t_{1}, 0) M(0, t_{2})$.

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91影视

Let \(X_{1}, X_{2}\) be two random variables with joint \(\operatorname{pdf} f\left(x_{1}, x_{2}\right)=x_{1} \exp \left\\{-x_{2}\right\\}\), for \(0

Short Answer

Step by step solution

Setting up the joint mgf

Substituting the pdf

Simplifying the integral

Solving the integral

Solving the remaining integral

Checking the equivalence

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