4

Suppose $X_1, X_2$ are independent random variables, with the same support $[0,1]$, on the same probability space with densities $f_1,f_2$ respectively.

By support I mean, $f_i$'s are $0$ outside $[0,1]$.

  1. We have that $$\int_{x} f_1(x)f_2(z-x) \,dx = \int_{x} g_1(x)g_2(z-x) \,dx$$ for all $z \in [0,2]$ and $g_i(y) = f_i(1-y)$ for both $i=\{1,2\}$.

  2. We also have that $$ f_i(x) \leq f_i(1-x), \forall x \in [0.5,1]$$ for both $i=\{1,2\}$.

I want to conclude that $$ f_1(x)f_2(z-x) = f_1(1-x)f_2(1-(z-x)) $$ for some $z$.

My try:

Rewrite (1) as $$\int_{x} (f_1(1-x)f_2(1-(z-x)) - f_1(x)f_2(z-x)) \,dx = 0. \quad (*)$$ For $z=1.5$, since we can restrict our interest to $0\leq z-x \leq 1$, we'll have both $x$ and $(z-x)$ exceeding $0.5$. Now from (2) we'll have $$ f_1(x) \leq f_1(1-x) ~\text{and} \\ f_2(z-x) \leq f_2(1-(z-x)).$$ So $$f_1(1-x)f_2(1-(z-x)) - f_1(x)f_2(z-x) \geq 0.$$ With $(*)$ we can in fact conclude $$ f_1(x)f_2(z-x) = f_1(1-x)f_2(1-(z-x)).$$

Now if the above conclusion holds can we say more? That is, from (2) I want to further conclude that $$ f_1(x) = f_1(1-x) $$ and $$f_2(z-x) =f_2(1-(z-x))$$ for some $z$.

Please comment on both the above conclusions I made. Thanks in advance for any help! Please feel free to make any further conclusions from these facts too, it'd be interesting to know them.

rookie
  • 1,738
  • What do you want independence for? – crankk Jul 20 '17 at 09:56
  • @crankk If you see point (1), the integral is nothing but the convolution of two densities. So having independence is convenient for me to find the PDF of the sum of random variables. So, can you comment on the two conclusions which I have posed? – rookie Jul 20 '17 at 10:02
  • 1
    I agree on your first conclusion. In fact $f_1(x_1)f_2(x_2)=f_1(1-x_1)f_2(1-x_2)$ holds for all $z \in [0,0.5]$ and $z \in [1.5,2]$. – crankk Jul 20 '17 at 10:27
  • @crankk Thank you for unmasking more information. About the second conclusion, if it's incorrect then can you provide a counterexample? – rookie Jul 20 '17 at 10:56
  • Sorry, I made a mistake. You get nothing from $(*)$ since you are integrating over a nullset. – crankk Jul 20 '17 at 11:32

1 Answers1

1

In (1) both sides are indeed $0$. the set $\{(x_1,x_2):x_1+x_2=z\}$ is a Nullset in $\mathbb{R}^2$ (w.r.t. Lebeguesmeasure), since it is the graph of the continuous function $x_1 \to z-x_1$. Therefore, (1) is meaningless and the only assumption left is $f_i(x) \le f_i(1-x)$ for $x \ge 0.5$. Now you can find counterexamples for both conclusions (f.e. $f_1(x)=f_2(x)=0.5x$ is a counterexample for conclusion 2).

crankk
  • 1,439
  • Let $X_1, X_2$ be uniform random variables in $[0,1]$. So the density of $Z= X_1+X_2$ will be $f_Z(z)=\int_{x_1,x_2:,x_1+x_2 = z} f_1(x_1)f_2(x_2) ,dx_1,dx_2 = \int_{-\infty}^\infty f_1(x)f_2(z-x),dx$. Finally you will get $$ f_Z(z) = \begin{cases} z & \text{for $0 < z < 1$} \ 2-z & \text{for $1 \le z < 2$} \ 0 & \text{otherwise.} \end{cases}$$ For more visit this : https://math.stackexchange.com/questions/357672/density-of-sum-of-two-uniform-random-variables-0-1 – rookie Jul 20 '17 at 11:51
  • I got the mistake. Am editing the question. – rookie Jul 20 '17 at 12:02
  • I was thinking of adding $X_1,X_2$ thus wrote generally as a double integral. Maybe that's a bad notation to start with. I have corrected the question now. Please take a look at it at your convenience. – rookie Jul 20 '17 at 12:13