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Problem 8.49 Let X and Y be independent random variables with densities fX(x) and fY(y), respectively. Show that the random variable Z = X+Y has a density given by \u222b \u221e\u2212 \u2212= z XYZ dssfszfzf )()()( Hint: [ ] ],[ XzYzXzZ \u2212\u2264\u2264=\u2264 PP Solution (Typo in problem statement - should be \u201cpositive\u201d independent random variables) Using the hint, we have that FZ(z) = P[Z \u2264 z] and \u222b \u222b \u221e\u2212 \u2212 \u221e\u2212 = z xz YXZ dxdyyfxfzF )()()( To differentiate this result with respect to z, we use the fact that if \u222b= b a dxzxhzg ),()( then dz dazah dz dbzbhdxzxh zdz zg b a ),(),(),()( \u2212+\u2202 \u2202=\u2202 \u222b (1) Inspecting FZ(z), we identify h(x,z) \u222b\u2212 \u221e\u2212 = xz YX dyyfxfzxh )()(),( and a = -\u221e and b = z. We then obtain \u222b \u222b \u222b\u222b \u222b\u222b \u221e\u2212 \u2212 \u221e\u2212 \u2212\u221e\u2212 \u221e\u2212\u221e\u2212 \u2212 \u221e\u2212 \u2212 \u221e\u2212 \u23a5\u23a6 \u23a4\u23a2\u23a3 \u23a1= \u22c5\u2212\u221e\u2212+\u23a5\u23a6 \u23a4\u23a2\u23a3 \u23a1= = z xz YX z YX z zz YX xz YX ZZ dxdyyfxf dz d dyyff dz dzdyyfzfdxdyyfxf dz d zF dz dzf )()( 0)()()()()()( )()( )( Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. page\u20268-60 Problem 8-49 continued where the second term of the second line is zero since the random variables are positive, and the third term is zero due to the factor zero. Applying the differentiation rule a second time, we obtain \u222b \u222b \u221e\u2212 \u221e\u2212 \u2212= \u23a5\u23a6 \u23a4\u23a2\u23a3 \u23a1 \u2212\u221e\u2212\u221e\u2212\u2212\u2212+= z YX z YXYXZ dxxzfxf dx dz dfxf dz xzdxzfxfzf )()( )()()()()()(0)( which is the desired result. An alternative solution is the following: we note that [ ] [ ] [ ] [ ] [ ]xzY zYx xXzYx xXzYXxXzZ \u2212\u2264= \u2264+= =\u2264+= =\u2264+==\u2264 P P P PP | || where the third equality follows from the independence of X and Y. By differentiating both sides with respect to z, we see that )()|(| xzfxzf YXZ \u2212= By the properties of conditional densities )()()|()(),( |, xzfxfxzfxfxzf YXXZXXZ \u2212== Integrating to form the marginal distribution, we have \u222b\u221e \u221e\u2212 \u2212= dxxzfxfzf YXZ )()()( If Y is a positive random variable then fY(z-x) is zero for x > z and the desired result follows. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. page\u20268-61