What is the conditional distribution function of P(T < z| U=x) where U, V are uniformly distributed on [0,1] and T= Max(U,V)

Question

Let U, V be independent random variables, both uniformly distributed on [0,1].T= max (U, V).

1) What is the joint distribution function of T and U, P(T <= Z, U <= x)?

I did the following:

P(T <= Z, U <= x)= P(max(U,V)<=z, U <= x) if x<=z, then the answer is xz. Because P(max(U,V)<=z, U <= x)= P(V <= z, U <=x) if x>z, then the answer is z^2, because P(max(U,V)<=z, U <= x)= P(V <=z, U <=z) Is this correct?

If so, should I add them together or should I separate the situation and say if x<=z, then the answer is xz and if x>z, then the answer is z^2?

2)derive the conditional distribution function P(max(U,V)<=z | U = x)?

I did the following : if x>z, the answer is 0 because it is impossible. if x<=z, P(max(U,V)<=z | U = x)= P(V <= z, U = x)/ P(U=x), since U ,V are independent, the answer is P (V <= z) = z

Is the correct?

3) Is this distribution function continuous for all z? Hint : first take the even x-1/n <= U <= x+ 1/n as the condition and use the usual formula for conditional probabilities, and then go to the limit as n goes infinity. But How to express P(max(U,V)<=z | x-1/n <= U <= x+ 1/n)?

Please can anyone help?

Answer 1

If so, should I add them together or should I separate the situation and say if x<=z, then the answer is xz and if x>z, then the answer is z^2?

The latter.   You have a piecewise function.   Indeed there are a few other cases to consider.

$$\begin{align}\mathsf P(U\leq x, T\leq z) ~=~& \begin{cases}\mathsf P(U\leq x, V\leq z) & : x<z \\[1ex] \mathsf P(U\leq z, V\leq z) & : z\leq x \end{cases} \\[1ex] =~& \begin{cases} 0 & : x< 0 ~\vee z<0 \\[1ex] xz & : 0\leq x < z\leq 1 \\[1ex] x & :0\leq x\leq 1<z \\[1ex] z^2 & : 0\leq z\leq x \leq 1~\vee~0\leq z\leq 1 <x \\[1ex]1 & : 1< z~\wedge~ 1<x \end{cases}\end{align}$$

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I did the following : if x>z, the answer is 0 because it is impossible. if x<=z, P(max(U,V)<=z | U = x)= P(V <= z, U = x)/ P(U=x), since U ,V are independent, the answer is P (V <= z) = z

So far so good.   Can you express it piecewise, as above?

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But How to express P(max(U,V)<=z | x-1/n <= U <= x+ 1/n)?

$$\lim_{n\to \infty} \mathsf P(T\leq z\mid x-1/n \leq U \leq x+ 1/n)~=~\lim_{n\to \infty} \dfrac{P(T\leq z, x-1/n\leq U\leq x+1/n)}{\mathsf P(x-1/n\leq U\leq x+1/n)}$$