We have given two PDFs as following:
$$f(x)=
\begin{cases}
3\lambda x^2 \exp(-\lambda x^3),& \text{if } x> 0\\
0, & \text{otherwise}
\end{cases}$$
$$g(y)=
\begin{cases}
3\mu y^2 \exp(-\mu y^3),& \text{if } y> 0\\
0, & \text{otherwise}
\end{cases}$$
Both of these PDFs are independent. What is the probability of $\lambda = \mu$ ie. $P(\lambda = \mu)$?
My attempt:
I have already shown that
$$P(X<Y) = \frac{\lambda}{\lambda+\mu}$$
Proof:
$$P(X<Y) = 1 – P(Y<X)$$
$$P(X<Y) = 1-\int_0^{\infty}\int_0^xf_{XY}(x,y)dy.dx$$
$$P(X<Y) = 1-\int_0^{\infty}\int_0^xf_{X}(x)f_{Y}(y)dy.dx$$
$$P(X<Y) = 1-\int_0^{\infty}f_{X}(x)\int_0^x3\mu y^2 \exp(-\mu y^3)dy.dx$$
$$P(X<Y) = 1+\int_0^{\infty}f_{X}(x)(\exp(-\mu x^3) – 1).dx$$
$$P(X<Y) = 1+\int_0^{\infty}3\lambda x^2 \exp(-\lambda x^3)(\exp(-\mu x^3) – 1).dx$$
$$P(X<Y) = 1-\int_0^{\infty}-3\lambda x^2 \exp(-(\lambda+\mu) x^3)dx + \int_0^{\infty}-3\lambda x^2 \exp(-\lambda x^3).dx$$
$$P(X<Y) = 1+\frac{\lambda}{\lambda+\mu} – 1$$
$$P(X<Y) = \frac{\lambda}{\lambda+\mu}$$
Best Answer
I read the original question from here. You indeed misread it. Here's what it is asking:
The first part is asking for $P(Y > X)$ which can be calculated as: \begin{align*} P(Y > X) &= \int_{0}^{\infty} \int_x^\infty \left( 3 \lambda x^2 \exp(-\lambda x^3) \right) \left( 3 \mu y^2 \exp(-\mu y^3) \right) dy \,dx\\ &= \int_0^\infty 3 \lambda x^2 \exp(-\lambda x^3) \left [\, \int_x^\infty 3 \mu \exp(-\mu y^3) dy \, \right ] dx \\ &= \frac{\lambda}{\lambda + \mu} \end{align*}
The second part is asking to evaluate the answer when $\lambda = \mu$ which is $\frac{1}{2}$.
Intuitively, $\lambda = \mu$ implies symmetry so $P(X > Y) = P(Y > X) = \frac{1}{2}$.
Notes: