We have a set of positive random variables $\boldsymbol X=\{X_1,X_2,\ldots\}$, where $X_1,X_2,\ldots,$ are independent and identically distributed (i.i.d.). The CDF $F(x)$ and PDF $f(x)$ for Xi are known in advance.

$K$ is an integer random variable with CDF $F_K(k)$ and PDF $f_K(k)$.

Define $S_n=\sum_{i=1}^n X_i$, then what is the distribution of $S_K$?


1 Answer 1


This would be easy if $K$ is constant (but it isnt't). This means that we don't know (yet) how to compute $P(S)$, but we know how to compute $P(S | K)$ ("if $K$ were a constant"). Well, then use conditional probabilities:

$$P(S) = \sum_K P(S,K) = \sum_K P(S|K) P(K)$$

  • $\begingroup$ Is there any simple way to compute $P(S\mid K)$? Because this involves $K$-fold convolution of $F(x)$ or $f(x)$. This is what I try to avoid. $\endgroup$
    – Bloodmoon
    Jul 4, 2015 at 11:24
  • 1
    $\begingroup$ No, I don't think you can avoid the convolution. Recall only that if you use CF (characteristic functions) it might be simpler, because convolutions translate to products. $\endgroup$
    – leonbloy
    Jul 4, 2015 at 11:36
  • $\begingroup$ I see, mathematically CF provides a way to avoid direct convolution calculation, but convert the CDF or PDF to Fourier Transform so that convolution can be translated to products, which is simpler. Then we can convert back by inverse Fourier Transform. Is this correct? (Of course CF also provides other functionalities.) $\endgroup$
    – Bloodmoon
    Jul 4, 2015 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.