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2d
answered Compactness and Hausdorffness with different topology
2d
revised Trouble with two equations with 4 unknowns
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2d
comment Trouble with two equations with 4 unknowns
I didn't mean $y=b,x=a$. I meant there are two solutions: #1 with $x=0, y=b$, #2 with $x=a, y=0$.
2d
answered Trouble with two equations with 4 unknowns
2d
comment Trouble with two equations with 4 unknowns
Including $x=0$, $y = b$ and $x=a$, $y=0$?
2d
comment Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
If the distribution of the $J$'s is unknown, you can't assert convergence for the right side of your original equation except at $t=0$. You don't even know that $E[Y_t^m]$ exists at all.
2d
answered How to come up with proofs of these results? Or, are these results true in the first place?
2d
comment Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
No, that's wrong. $E[Y_t^m] = \exp(\lambda t (E[J_1^m] - 1))$, not $\exp(\lambda t (E[J_1]^m-1))$.
2d
comment Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
The only way you could have $E[J_k^m] = c^m$ (and thus $E[Y_t^m] = e^{\lambda t (c^m-1)}$) for all $m$ is if $J_k = c$ with probability $1$.
2d
comment Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
... and what is the distribution of the $J$'s? You'll need to know about $E[J^m]$, not just $E[J]$, to get $E[Y^m]$.
2d
revised Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
added 562 characters in body
2d
revised Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
deleted 76 characters in body
2d
revised Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
deleted 76 characters in body
2d
answered Interchange of the expected value and infinite summation $E(\sum_{m=0}^\infty (it)^m Y_t^m/m!)=\sum_{m=0}^\infty E((it)^m Y_t^m/m!)$
2d
comment I have a trouble with an integration from a book of kinetic gas theory.
en.wikipedia.org/wiki/Exponential_growth#Differential_equation
2d
answered Three Simultaneously Diagonalizable Matrices
2d
answered Combinations with maximum allowed Repetition
Apr
23
answered Exchangeable Random Variable but not independent?
Apr
23
answered Is every subset of a Borel set Lebesgue measurable?
Apr
23
comment Largest age difference between great-great-…-grandparents?
But even in your model, any of the $Y$'s that share a common descendant before you will be dependent. For example, the $Y$ for your mother's father and your mother's mother are dependent because they share the $X$ for your mother.