A small circuit with two diodes
An ideal diode analysis gives you an approximate answer:


The voltage across the lamp is approximately 9.3-9.5 V.
The diode carrying current I is reverse biased.


For a real diode, there will be some reverse leakage current through the "I" diode. This means I could be anywhere from -1 pA to -1 uA, depending on the type of diode. If you choose something like a Shottky diode to minimize the drop when the diode is forward biased, you will probably increase the amount of reverse leakage when it is reverse biased.

Final answer, in a realistic scenario \$\frac{I}{J}\$ will be a small but non-zero negative number.