What is the new pdf after making the transformation?

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My question is from George Casella statistical inference textbook Question 8.9 (b). I know how to do this question and I worked out. But I don't agree with the solution. In the question, it said "make the transformation $X_i = 1/Y_i$. Then I just replace $Y_i$ with $1/X_i$ in the original pdf $\lambda_i e^{-\lambda_i y_i}$. So my new pdf is just $\lambda_i e^{-\lambda_i/x_i}$.

However, the solution said the new pdf is $(\lambda_i/x_i^2) e^{-\lambda_i/x_i}$.

Does the solution have the typo? I used my pdf and I can solve out the problem.

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The explanation forgets to include the Jacobian term. When in doubt, always go back to the explicit formula for performing change of variables. Assuming we have a transformation $X=g(Y)$ that is monotone on the support of $Y$ we can use the formula $$ f_X(x)=f_Y(g^{-1}(x))|J|, $$ where $|J|$ is the Jacobian of the transformation. In this particular case we have $X=1/Y$ (which is monotone on $\operatorname{supp}(Y)=(0,\infty)$) and so $g^{-1}(x)=1/x$ and $|J|=|\partial_x 1/x|=1/x^2$. Hence, $$ f_X(x)=f_Y(1/x)/x^2=(\lambda/x^2)\exp(-\lambda/x),\quad x>0, $$ which is the correct solution as given by the text.