Conditional probability given V=3

Question

X and Y are independent exponential random variables with mean 1. If V = X+Y find the conditional density of X given V = 3

Therefore $$f(x) = e^{-x}, x>0$$ and $$f(y) = e^{-y},y>0$$

To get the pdf of V, I'm using the convolution formula $$fv(v) = \int_0^v fx(x)fy(v-x) dx = ve^{-v}$$

Here is my question. For the conditional probability $$f(X|V=3) = \frac{f(X|V=3)}{f(V=3)}$$ Given that $$f(V=3) = 3e^{-3}$$

How can i find? $$f(X|V=3)$$

And is this the right approach or am i making a mistake somewhere?

Answer 1

Your numerator is incorrect: $$f_{X \mid V}(x \mid 3) = \dfrac{f_{X, V}(x, 3)}{f_{V}(3)}$$