MAP estimator of Z=X/Y estimating X

Question

The probabilities are given $$pdf(y)=\lambda_ye^{-\lambda_y y}, y\geq0$$ $$pdf(x)=\lambda_xe^{-\lambda_x x}, x\geq0$$ I have the observation such that $z=x/y$. I want to estimate $x$. First, I found $$pdf(z)={\lambda_x\lambda_y\over{(\lambda_x {z}}+\lambda_y)^2}$$. I know that MAP estimator of $x$ is $$\hat x=arg[max_{\theta}(Pr(x|z))]$$ $$Pr(x|z)={pdf(z|x)pdf(x)\over pdf(z)}$$ So, $$pdf(z|x)={\lambda_x\lambda_y\over{(\lambda_x {x \over y}}+\lambda_y)^2}$$ right? So I just maximize $$pdf(z|x)pdf(x)$$ right? Sorry for notation

Answer 1

I am not quite sure whether the following equation holds: $$pdf(z|x) = \frac{\lambda_x \lambda_y}{(\lambda_x \frac{x}{y} + \lambda_y)^2}$$

It seems that: $$\begin{align} pdf_Z(z|x)dz &= P(Z = z|X = x) \\ &= P(\frac{X}{Y} = z|X=x) \\ &=P(Y = \frac{x}{z}|X = x) \end{align}$$

According to your pdf of Z, it seems that X and Y are assumed independent, so: $$\begin{align} pdf(z|x)dz &= P(Y = \frac{x}{z}|X = x) \\ &= P(Y = \frac{x}{z})\\ &= pdf_Y(\frac{x}{z})dy \\ &= pdf_Y(\frac{x}{z})(\bigl\lvert-\frac{x}{z^2}\bigr\rvert dz)\\ \\ pdf(z|x)&=\frac{x}{z^2}\lambda_2e^{-\lambda_2\frac{x}{z}} \end{align}$$