Question

The random variables \(X\) and \(Y\) have the following properties: \(X\) is positive, i.e., \(P\\{X>0\\}=1\), with continuous density funetion \(f(x)\), and \(Y \mid X\) has a uniform distribution on \(\\{0, X\\} .\) Prove: If \(Y\) and \(X-Y\) are independently dis* tributed, then $$ f(x)=a^{2} x e^{-a x}, \quad x>0, \quad a>0 $$

Short answer

In order to prove that the density function \(f(x)=a^2 xe^{-ax}\), where \(x>0\) and \(a>0\), we first find the joint density function of Y and X and then the marginal density functions of Y and X-Y. We use the fact that Y and X-Y are independent, so their joint density function is the product of their marginal density functions. By comparing the resulting expressions, we obtain the desired density function \(f(x)=a^2 xe^{-ax}\).

Step-by-step solution

Find the joint conditional density function of Y and X given X

Since Y given X is uniformly distributed over the interval [0, X], we can write its probability density function as: \[f_{Y|X}(y|x) = \frac{1}{x}\]

Find the joint density function of Y and X

Using the definition of conditional density, we have: \[f_{Y,X}(y,x) = f_{Y|X}(y|x)f_X(x) = \frac{1}{x}f_X(x) \]

Find the marginal density functions of Y and X-Y

To find the marginal density functions, we need to integrate the joint density function with respect to the other variable. For f_Y(y), we integrate over x, and for f_(X-Y)(x-y), we integrate over y. \[f_Y(y) = \int_0^{\infty} \frac{1}{x}f_X(x) dx\] \[f_(X-Y)(x-y) = \int_0^{\infty} \frac{1}{y}f_X(y) dy\]

Use the independence of Y and X-Y

Since Y and X-Y are independent, their joint density function is the product of their marginal density functions: \[f_{Y,X}(y,x) = f_Y(y)f_(X-Y)(x-y)\] Substituting the expressions for the marginal density functions from Step 3, we have: \[\frac{1}{x}f_X(x) = \left(\int_0^{\infty} \frac{1}{x}f_X(x) dx\right)\left(\int_0^{\infty} \frac{1}{y}f_X(y) dy\right)\]

Derive the form of f(x)

We notice that both sides of the last equation have \(\int_0^{\infty} \frac{1}{x}f_X(x) dx\) as a common factor. Hence, we can write: \[\frac{1}{x}f_X(x) = C \int_0^{\infty} \frac{1}{y}f_X(y) dy\] Now we differentiate both sides with respect to x, using Leibniz's rule on the right-hand side: \[f'(x) - \frac{1}{x}f(x) = -C \frac{1}{x^2}f(x)\] Rearranging the terms, we get: \[\frac{f'(x)}{f(x)} = \left(\frac{1}{x} - C\frac{1}{x^2}\right)f(x)\] Integrating both sides with respect to x, we have: \[\ln (f(x)) = -ax + b\] By exponentiating both sides, we obtain \[f(x) = e^{-ax + b}\] Now, by comparing this expression to the desired form, we can see that the desired density function can be obtained by setting: \[a^2 x = e^b\] Which finally gives us the desired density function as: \[f(x) = a^2 xe^{-ax}\] This completes the proof.