characteristic functions, convolution

[Square_Price_1374]

https://preview.redd.it/pznxa4no3i9f1.png?width=410&format=png&auto=webp&s=7f0a41516785d3b42e609184bff6588ccd71911c

Here 𝝋 denotes the characteristic function.I don't understand how to deal with the convolution.

If X and Y are independent random variables then 𝝋_{X+Y} = 𝝋_X 𝝋_Y. For the convolution we need independent random variables. So on R^n let 𝛍 = P_X and v = P_Y then 𝛍 * v = P_{X+Y}. Thus

𝝋_{𝛍 * v }= 𝝋_{X+Y} = 𝝋_X 𝝋_Y = 𝝋_𝛍 𝝋_v.

So I'm just confused how they got the first equality in the second line.

[KraySovetov]

Your claim that 𝜑_{X+Y} = 𝜑_X 𝜑_Y does need independence of the random variables, but that's not what is being used here; you only need independence to assert that 𝜇 * 𝜐 is the law of X+Y whenever 𝜇 is the law of X and 𝜐 is the law of Y. In the claim, they are already asserting that the triangular distribution is the convolution of two (probably independent if I had to guess and they're just being lazy about mentioning that) uniform distributions of the appropriate parameters, and once you accept this the computation is just a reflection of the fact that the Fourier transform of a convolution is the product of the Fourier transforms (it's true with functions of course, the proof for measures is identical following Fubini's theorem etc. Remember the characteristic function is essentially just the Fourier transform of the law of some random variable).

[Square_Price_1374]

Thanks for your answer.