Product of independent random variables and tail deconvolution

Suppose $ X, Y$ are two independent non-negative random variables. The conditions

(i) $ \mathbb{P}(X > t) = \frac{C}{t^p} + o(t^{-p})$

(ii) $ \mathbb{P}(Y > t) = o(t^{-q})$ for any $ q > 0$

imply

(iii) $ \mathbb{P}(XY > t) = \frac{C \mathbb{E}[Y^p]}{t^p} + o(t^{-p})$ .

(Of course here I am talking about the asymptotic behaviour as $ t \to \infty$ and $ p > 0$ .)

My question concerns a converse of this statement: if I know (ii) and (iii), does that imply (i)?

(While I would very much love to see that this is true, I have the impression that this claim is false but just haven’t come up with a counter-example.)

I am aware that the converse holds at the exponential level, i.e. $ $ \lim_{t \to \infty} \frac{\log \mathbb{P}(X > t)}{\log t} = -p.$ $

One may consider random variable $ Y$ with a density (which is sufficient for my purpose) if that helps. In case a counter-example for the “full” converse can be found, I would like to know if the “full” converse can still hold when (a) $ Y$ is a lognormal random variable or slightly more generally (b) $ Y$ has a tail upper bounded by that of some lognormal.

Update: to clarify, $ Y$ is a given random variable, the distribution of which is hence given and cannot be chosen freely. In particular $ Y$ is not a constant (otherwise the converse is trivially true unless $ Y = 0$ a.s., in which case the converse is trivially false).

Deconvolution using Fourier transforms

I have a 2D signal in the form of a function $ g(x_m,y_m)$ given as

$ $ \begin{aligned} g(x_m,y_m) = \ & \int^{\infty}_{-\infty} \mathrm{d}x_o \ \mathrm{d}y_o \ \frac{1}{\varepsilon^2} P(x_o – x_m,y_o -y_m) \ & \quad \times \int^{\infty}_{-\infty} \mathrm{d}x_i \ \mathrm{d}y_i \ R(x_o – x_i,y_o -y_i)L(x_i,y_i) \end{aligned} \tag{1}$ $

The integrals in equation $ (1)$ can be seen as a convolution of $ P,R$ and $ L$ as $ $ g(x_m,y_m)= ((P/\varepsilon^2)*R*L)(x_m,y_m) \tag{2}$ $

I want to find $ L$ when $ g$ ,$ R$ and $ P$ are given.

I tried to use Fourier transforms to find $ L$ as:

$ $ L = \mathcal{F}^{-1} \left[ \frac{\mathcal{F}[g]}{\mathcal{F}[R] \cdot \mathcal{F}[P/\varepsilon^2]} \right]$ $

R[x_, y_] := 0.609739 E^(-444.116 (-0.00244704 + x)^2 - 444.116 (-0.0322566 + y)^2) +  0.0691803 E^(-48.9858 (-0.0105298 + x)^2 -  48.9858 (0.0054951 + y)^2) +  0.66442 E^(-426.449 (0.00315949 + x)^2 -  426.449 (0.0248433 + y)^2);   g[x_, y_] := 3.06909 E^(-18.585 x^2 + x (13.6144 - 27.7795 y) + (12.2542 - 18.1432 y) y) +  0.402245 E^(-7.37814 x^2 + x (9.61481 - 5.57202 y) + (6.46554 - 6.35048 y) y) + 120.468 E^(-0.0245919 x^2 + x (-0.00325668 + 0.00197362 y) + (-0.00103919 - 0.0281421 y) y) + 0.773818 E^(-3.79704 x^2 + (-15.0351 - 16.3606 y) y + x (1.75472 + 2.86666 y)) + 0.0833316 E^(-38.7396 x^2 + (-10.6949 - 14.1731 y) y + x (32.7513 + 18.7984 y));  epsilon = 0.048; P[x_, y_] := E^(-(2/epsilon)^2 (x^2 + y^2));  FTR = FourierTransform[R[x, y], {x, y}, {u, v}, FourierParameters -> {0, -2 \[Pi]}]; FTg = FourierTransform[g[x, y], {x, y}, {u, v}, FourierParameters -> {0, -2 \[Pi]}]; FTP = FourierTransform[P[x, y], {x, y}, {u, v}, FourierParameters -> {0, -2 \[Pi]}];  InverseFourierTransform[FTg/((1/epsilon^2)FTP*FTR),{u, v}, {xi, yi}, FourierParameters -> {1,-2*Pi}] 

However, the InverseFourierTransform takes a lot of time and does not return any result. How do I find $ L(x_i,y_i)$ ?, am I doing something wrong here?.