Example 4.18. Let X and Y be i.i.d. - N (0, o²), so that fxr(x, y) = fx(x)fy(y) 1 1 1 for all æ and y. Let Z = X² + Y². Then fz(z)dz fxy(x, y)dædy {(1,y):/a VI²+y²€[z,z+dz)} 27 z+dz fxy(r cos 0, r sin...


Work out Example 4.18 in the lecture notes by using the CDF method.


Example 4.18. Let X and Y be i.i.d. -<br>N (0, o²), so that<br>fxr(x, y) = fx(x)fy(y)<br>1<br>1<br>1<br>for all æ and y. Let Z = X² + Y². Then<br>fz(z)dz<br>fxy(x, y)dædy<br>{(1,y):/a<br>VI²+y²€[z,z+dz)}<br>27<br>z+dz<br>fxy(r cos 0, r sin 0)rdrd® (cf. Figure 4.8)<br>-2T<br>z+dz<br>„² cos² 0+r² sin² ,<br>´rdrd0<br>e<br>1<br>r2m<br>z+dz<br>rdrd0<br>27o?<br>-2T<br>1<br>zdzd0<br>2no?<br>-27<br>1<br>do<br>2no2<br>zpz -<br>1<br>e¯20* zdz(27)<br>* zpZ -<br>Cancelling dz on both sides, we have<br>1<br>fz(z) =<br>for all z > 0. Obviously, fz(z) = 0 for all z < 0.<br>The reader is encouraged to repeat the above example by the CDF method.<br>

Extracted text: Example 4.18. Let X and Y be i.i.d. - N (0, o²), so that fxr(x, y) = fx(x)fy(y) 1 1 1 for all æ and y. Let Z = X² + Y². Then fz(z)dz fxy(x, y)dædy {(1,y):/a VI²+y²€[z,z+dz)} 27 z+dz fxy(r cos 0, r sin 0)rdrd® (cf. Figure 4.8) -2T z+dz „² cos² 0+r² sin² , ´rdrd0 e 1 r2m z+dz rdrd0 27o? -2T 1 zdzd0 2no? -27 1 do 2no2 zpz - 1 e¯20* zdz(27) * zpZ - Cancelling dz on both sides, we have 1 fz(z) = for all z > 0. Obviously, fz(z) = 0 for all z < 0.="" the="" reader="" is="" encouraged="" to="" repeat="" the="" above="" example="" by="" the="" cdf="">

Jun 05, 2022
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