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Chapter 2: Problem 4

Let \(X_{1}, X_{2}, X_{3}\) be iid with common pdf \(f(x)=e^{-x}, x>0,0\) elsewhere. Find the joint pdf of \(Y_{1}=X_{1}, Y_{2}=X_{1}+X_{2}\), and \(Y_{3}=X_{1}+X_{2}+X_{3}\).

Short Answer

Expert verified
The joint pdf of \(Y_{1}=X_{1}, Y_{2}=X_{1}+X_{2}\), and \(Y_{3}=X_{1}+X_{2}+X_{3}\) is \(f_{Y_{1}, Y_{2}, Y_{3}}(y_{1}, y_{2}, y_{3}) = e^{-y3}\), when \(0 < y1 < y2 < y3\), and 0 elsewhere.

Step by step solution

01

Understanding the Change of Variable

We need to change variables as per \(Y_{1}=X_{1}, Y_{2}=X_{1}+X_{2}\), and \(Y_{3}=X_{1}+X_{2}+X_{3}\). So, for that, you need to rewrite these variables in terms of \(Y_{1}, Y_{2}, Y_{3}\). They become: \(X_{1}=Y_{1}, X_{2}=Y_{2}-Y_{1}, X_{3}=Y_{3}-Y_{2}\).
02

Find the Jacobian

We also need the Jacobian of the transformation, which is \(|J|=\left|\frac{\partial(X_{1}, X_{2}, X_{3})}{\partial(Y_{1}, Y_{2}, Y_{3})}\right|\). This can be found out from the determinant of the matrix. The Jacobian is equal to 1.
03

Applying the Transformation

Given that \(X_{1}, X_{2}, X_{3}\) are iid, we can say \(f_{X_{1}, X_{2}, X_{3}}(x_{1}, x_{2}, x_{3}) = f_{X_{1}}(x_{1})f_{X_{2}}(x_{2})f_{X_{3}}(x_{3})\). Now, we have to substitute \(x_{1}=y_{1}, x_{2}=y_{2}-y_{1}, x_{3}=y_{3}-y_{2}\) into the joint pdf \(f_{X_{1}, X_{2}, X_{3}}(x_{1}, x_{2}, x_{3})\). The result is \(f_{Y_{1}, Y_{2}, Y_{3}}(y_{1}, y_{2}, y_{3}) = e^{-y1}e^{-y2+y1}e^{-y3+y2} = e^{-y3}\), for \(0 < y1 < y2 < y3\), and 0 elsewhere.

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Most popular questions from this chapter

Given that the nonnegative function \(g(x)\) has the property that $$ \int_{0}^{\infty} g(x) d x=1 $$ show that $$ f\left(x_{1}, x_{2}\right)=\frac{2 g\left(\sqrt{x_{1}^{2}+x_{2}^{2}}\right)}{\pi \sqrt{x_{1}^{2}+x_{2}^{2}}}, 0

If \(p\left(x_{1}, x_{2}\right)=\left(\frac{2}{3}\right)^{x_{1}+x_{2}}\left(\frac{1}{3}\right)^{2-x_{1}-x_{2}},\left(x_{1}, x_{2}\right)=(0,0),(0,1),(1,0),(1,1)\), zero elsewhere, is the joint pmf of \(X_{1}\) and \(X_{2}\), find the joint \(\mathrm{pmf}\) of \(Y_{1}=X_{1}-X_{2}\) and \(Y_{2}=X_{1}+X_{2}\)

Let \(X_{1}, X_{2}\) be two random variables with joint \(\operatorname{pmf} p\left(x_{1}, x_{2}\right)=(1 / 2)^{x_{1}+x_{2}}\), for \(1 \leq x_{i}<\infty, i=1,2\), where \(x_{1}\) and \(x_{2}\) are integers, zero elsewhere. Determine the joint mgf of \(X_{1}, X_{2} .\) Show that \(M\left(t_{1}, t_{2}\right)=M\left(t_{1}, 0\right) M\left(0, t_{2}\right)\).

Let the random variables \(X_{1}\) and \(X_{2}\) have the joint pmf described as follows: $$ \begin{array}{c|cccccc} \left(x_{1}, x_{2}\right) & (0,0) & (0,1) & (0,2) & (1,0) & (1,1) & (1,2) \\ \hline p\left(x_{1}, x_{2}\right) & \frac{2}{12} & \frac{3}{12} & \frac{2}{12} & \frac{2}{12} & \frac{2}{12} & \frac{1}{12} \end{array} $$ and \(p\left(x_{1}, x_{2}\right)\) is equal to zero elsewhere. (a) Write these probabilities in a rectangular array as in Example 2.1.3, recording each marginal pdf in the "margins." (b) What is \(P\left(X_{1}+X_{2}=1\right)\) ?

Five cards are drawn at random and without replacement from an ordinary deck of cards. Let \(X_{1}\) and \(X_{2}\) denote, respectively, the number of spades and the number of hearts that appear in the five cards. (a) Determine the joint pmf of \(X_{1}\) and \(X_{2}\). (b) Find the two marginal pmfs. (c) What is the conditional pmf of \(X_{2}\), given \(X_{1}=x_{1}\) ?
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